Id of bills with the highest number of distinct items

Select BillID From L.BillItem
Group By BillID Having Count(*) >= ALL
(Select Count(*) From L.BillItem Group By BillID)

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Very hard: List all couples of cinema (names only), that have screened same movies (the same set of movies).
List only couples, where Name1<Name2.

select C1.Name as Name1, C2.Name as Name2
from Cinemas C1
inner join Cinemas C2 -- Couples of Cinemas ...
on (C1.Name < C2.Name) -- ... ordered
-- not to write couple of same cinemas
-- nor one couple twice ...
where not exists( -- ... where doesn't exist ...
select *
from Program P1 -- ... a movie, screened ...
where P1.Name=C1.Name -- ... by first cinema ...
and not exists( -- ... such that wouldn't exist ...
select *
from Program P2 -- ... record, ...
where P2.Name=C2.Name -- ... that the movie was screened ...
and P2.Title=P1.Title -- ... in the second cinema as well
)
)
and not exists( -- and vice versa
select *
from Program P2
where P2.Name=C2.Name
and not exists(
select *
from Program P1
where P1.Name=C1.Name
and P1.Title=P2.Title
)
);

---

List all cinemas, that have screened exactly three movies from Cameron (movies, not screenings) - all columns

select *
from Cinemas C
where 3 = (
select count(distinct M.Title) -- important is the distinct
from Program P
inner join Movies M
on (M.Title=P.Title)
where P.Name=C.Name
and M.Director='Cameron'
);

---

Id of bills with the highest number of distinct items

Select BillID From L.BillItem
Group By BillID Having Count(*) >= ALL
(Select Count(*) From L.BillItem Group By BillID)

---

List names of cinemas with number of their screenings. There are not cinemas without any screening.

select program.Name, count(*) from Program program
group by program.Name

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ID of bills with the most expensive item in the shop

select distinct bi.BillID from BillItem bi
inner join Item item on item.ID = bi.ItemID
where item.Price >= (select max(Price) from Item)

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