diff --git a/sql/task1.sql b/sql/task1.sql
index 90de336ca..56e4148cb 100644
--- a/sql/task1.sql
+++ b/sql/task1.sql
@@ -1,14 +1,38 @@
 -- Problem 1: Retrieve all products in the Sports category
 -- Write an SQL query to retrieve all products in a specific category.
 
+SELECT *  --selecting all columns
+FROM Products --from the Products table
+WHERE category_id = 1; --where the category_id is 1 (Sports)
+
+
+
+
 -- Problem 2: Retrieve the total number of orders for each user
 -- Write an SQL query to retrieve the total number of orders for each user.
 -- The result should include the user ID, username, and the total number of orders.
 
+SELECT u.user_id, u.username, COUNT(o.order_id) AS total_orders --selecting the user_id, username, and total number of orders
+FROM Users AS u  -- from the Users table
+LEFT JOIN Orders AS o ON u.user_id = o.user_id --left joining the Orders table on the user_id
+GROUP BY u.user_id, u.username; --grouping by the user_id and username
+
 -- Problem 3: Retrieve the average rating for each product
 -- Write an SQL query to retrieve the average rating for each product.
 -- The result should include the product ID, product name, and the average rating.
 
+SELECT p.product_id, p.product_name, AVG(r.rating) AS average_rating --selecting the product_id, product_name, and average rating
+FROM Products AS p --from the Products table
+LEFT JOIN Reviews AS r ON p.product_id = r.product_id --left joining the Reviews table on the product_id
+ORDER BY p.product_id; --ordering by the product_id
+
 -- Problem 4: Retrieve the top 5 users with the highest total amount spent on orders
 -- Write an SQL query to retrieve the top 5 users with the highest total amount spent on orders.
 -- The result should include the user ID, username, and the total amount spent.
+
+SELECT u.user_id, u.username, SUM(o.total_amount) AS total_amount_spent --selecting the user_id, username, and total amount spent
+FROM Users AS u --from the Users table
+JOIN Orders AS o ON u.user_id = o.user_id -- joining the Orders table on the user_id
+GROUP BY u.user_id, u.username -- grouping by the user_id and username
+ORDER BY total_amount_spent DESC -- ordering by the total amount spent in descending order
+LIMIT 5; -- limiting the results to 5
diff --git a/sql/task2.sql b/sql/task2.sql
index ad2596731..1c4dca6f2 100644
--- a/sql/task2.sql
+++ b/sql/task2.sql
@@ -3,17 +3,93 @@
 -- The result should include the product ID, product name, and the average rating.
 -- Hint: You may need to use subqueries or common table expressions (CTEs) to solve this problem.
 
+-- Create a CTE to calculate the average rating for each product
+-- Create a CTE to rank the products by average rating
+WITH ProductRatings AS (
+    SELECT
+        product_id,
+        AVG(rating) AS average_rating
+    FROM Reviews
+    GROUP BY product_id
+)
+
+
+, RankedProducts AS ( 
+            
+    SELECT
+        p.product_id,
+        p.product_name,
+        pr.average_rating,
+        RANK() OVER (ORDER BY pr.average_rating DESC) AS rating_rank
+    FROM Products AS p
+    JOIN ProductRatings AS pr ON p.product_id = pr.product_id
+)
+
+
+SELECT -- Retrieve the products with the highest average rating
+    product_id,
+    product_name,
+    average_rating
+FROM RankedProducts
+WHERE rating_rank = 1;
+
+
 -- Problem 6: Retrieve the users who have made at least one order in each category
 -- Write an SQL query to retrieve the users who have made at least one order in each category.
 -- The result should include the user ID and username.
 -- Hint: You may need to use subqueries or joins to solve this problem.
 
+-- Select users who have ordered from all categories
+SELECT 
+    u.user_id, 
+    u.username
+FROM 
+    Users AS u
+WHERE 
+    -- Count the distinct categories from which the user has ordered
+    (SELECT COUNT(DISTINCT p.category_id)
+     FROM Orders AS o
+     JOIN Order_Items AS oi ON o.order_id = oi.order_id
+     JOIN Products AS p ON oi.product_id = p.product_id
+     WHERE o.user_id = u.user_id) =
+    -- Compare to the total number of categories available
+    (SELECT COUNT(*) FROM Categories);
+
+
 -- Problem 7: Retrieve the products that have not received any reviews
 -- Write an SQL query to retrieve the products that have not received any reviews.
 -- The result should include the product ID and product name.
 -- Hint: You may need to use subqueries or left joins to solve this problem.
 
+-- Select products that have not received any reviews
+SELECT p.product_id, p.product_name
+FROM Products AS p
+LEFT JOIN Reviews AS r ON p.product_id = r.product_id
+WHERE r.product_id IS NULL;
+
 -- Problem 8: Retrieve the users who have made consecutive orders on consecutive days
 -- Write an SQL query to retrieve the users who have made consecutive orders on consecutive days.
 -- The result should include the user ID and username.
--- Hint: You may need to use subqueries or window functions to solve this problem.
\ No newline at end of file
+-- Hint: You may need to use subqueries or window functions to solve this problem.
+
+-- Create a CTE to retrieve the users who have made orders on consecutive days
+WITH OrderedUsers AS ( 
+    SELECT
+        u.user_id,
+        u.username,
+        o.order_date,
+        LAG(o.order_date) OVER (PARTITION BY o.user_id ORDER BY o.order_date) AS previous_order_date
+    FROM
+        Orders AS o
+    JOIN Users AS u ON o.user_id = u.user_id
+)
+ 
+SELECT DISTINCT -- Select users who have made consecutive orders on consecutive days
+    user_id,
+    username
+FROM
+    OrderedUsers
+WHERE
+    order_date = DATE(previous_order_date, '+1 day');
+
+
diff --git a/sql/task3.sql b/sql/task3.sql
index f078a9439..0c1709498 100644
--- a/sql/task3.sql
+++ b/sql/task3.sql
@@ -3,17 +3,68 @@
 -- The result should include the category ID, category name, and the total sales amount.
 -- Hint: You may need to use subqueries, joins, and aggregate functions to solve this problem.
 
+-- Select category details and calculate total sales amount
+SELECT c.category_id, c.category_name, SUM(oi.quantity * oi.unit_price) AS total_sales_amount
+FROM Categories c
+JOIN Products p ON c.category_id = p.category_id  -- Link products to their categories
+JOIN Order_Items oi ON p.product_id = oi.product_id  -- Get sales data for products
+GROUP BY c.category_id, c.category_name  -- Aggregate sales by category
+ORDER BY total_sales_amount DESC  -- Order categories by total sales
+LIMIT 3;  -- Return only the top 3 categories
+
+
 -- Problem 10: Retrieve the users who have placed orders for all products in the Toys & Games
 -- Write an SQL query to retrieve the users who have placed orders for all products in the Toys & Games
 -- The result should include the user ID and username.
 -- Hint: You may need to use subqueries, joins, and aggregate functions to solve this problem.
 
+-- Find users who ordered all 'Toys & Games' products
+SELECT u.user_id, u.username
+FROM Users u
+JOIN Orders o ON u.user_id = o.user_id  -- Join orders to users
+JOIN Order_Items oi ON o.order_id = oi.order_id  -- Get items in those orders
+JOIN Products p ON oi.product_id = p.product_id  -- Link those items to products
+WHERE p.category_id = (
+  SELECT category_id FROM Categories WHERE category_name = 'Toys & Games'  -- Target 'Toys & Games' category
+)
+GROUP BY u.user_id, u.username  -- Group by user, Ensure user ordered all products in category
+HAVING COUNT(DISTINCT p.product_id) = (  
+  SELECT COUNT(*) FROM Products WHERE category_id = (
+    SELECT category_id FROM Categories WHERE category_name = 'Toys & Games'
+  )
+);
+
+
 -- Problem 11: Retrieve the products that have the highest price within each category
 -- Write an SQL query to retrieve the products that have the highest price within each category.
 -- The result should include the product ID, product name, category ID, and price.
 -- Hint: You may need to use subqueries, joins, and window functions to solve this problem.
 
+-- Rank products by price within each category
+WITH RankedProducts AS (
+  SELECT product_id, product_name, category_id, price,
+  RANK() OVER (PARTITION BY category_id ORDER BY price DESC) AS price_rank
+  FROM Products
+)
+
+
+SELECT product_id, product_name, category_id, price -- Select highest-priced products in each category
+FROM RankedProducts
+WHERE price_rank = 1;
+
+
 -- Problem 12: Retrieve the users who have placed orders on consecutive days for at least 3 days
 -- Write an SQL query to retrieve the users who have placed orders on consecutive days for at least 3 days.
 -- The result should include the user ID and username.
 -- Hint: You may need to use subqueries, joins, and window functions to solve this problem.
+-- ee
+
+-- Identify consecutive order days for each user
+WITH consecutivedays AS (
+  SELECT user_id, order_date, DATE(order_date) - ROW_NUMBER() OVER (PARTITION BY user_id ORDER BY order_date) AS grp FROM Orders
+),
+-- Count the length of each consecutive day streak
+consec AS (
+  SELECT user_id, COUNT(*) AS streak_length FROM consecutivedays GROUP BY user_id, grp HAVING COUNT(*) >= 3
+)
+SELECT DISTINCT o.user_id, u.username FROM consec o JOIN Users u ON o.user_id = u.user_id;