1

USE Store
SELECT Category.name, COUNT(Product.id) AS product_count
FROM Product

INNER JOIN Category
ON Product.id_category = Category.id

INNER JOIN Delivery
ON Product.id = Delivery.id_product

GROUP BY Category.name
HAVING AVG(Delivery.price) > 100

2

USE Store
SELECT Category.name, product.name, SUM(Sale.price) AS price
FROM Product

INNER JOIN Category
ON Product.id_category = Category.id

INNER JOIN Sale
ON Product.id = Sale.id_product
WHERE Category.name = 'Фрукти' OR Category.name = 'Цукерки'

GROUP BY Category.name, Product.name

4

USE Store
SELECT Category.name, MIN(Product.quantity) AS quantity
FROM Product

INNER JOIN Category
ON Product.id_category = Category.id

GROUP BY Category.name
ORDER BY MIN(Product.quantity) 

--прогуглив та знайшов для повного виконання завдання максимум таке: SELECT VALUE p FROM AdventureWorksEntities.Products AS p ORDER BY p.ListPrice LIMIT(@limit)
--але такий синтаксис чомусь не спрацював

5

USE Store
SELECT Category.name, Product.quantity
FROM Product

INNER JOIN Category
ON Product.id_category = Category.id

INNER JOIN Delivery
ON Delivery.id_product = Product.id

INNER JOIN Supplier
ON Delivery.id_supplier = Supplier.id

WHERE Delivery.price > 400
AND Supplier.id = 1 OR Supplier.id = 2 OR Supplier.id = 3