First a remark :
It is better not to use the expression “ negation” when one talks about a set; a set cannot be negated, only a sentence can be negated; but one can say that the set called “complement of the set A ∩ B” is defined by the negation of the sentence defining A ∩ B. That is, if the symbol for the complement of A ∩ B is : (A ∩ B)’ then one can say that :
(A ∩ B)’ = the set of all x such that it is false that (x belongs to A & x belongs to B).
The complement of a set S , denotetd by the symbol : S' , is the set of all x ( belonging by definition to the universal set U) that do not belong to B, that is the set : U – S.
Now regarding your question :
If the set A also is your universal set U , that is, if U=A
then , in that special case :
A-B = U-B = the set of all x that do not belong to B= B’ = complement of B
(A ∩ B)’ = U – (A ∩ B) = U – (U∩ B) = U – B = B’ = complement of B.
and therefore ( still in the special case we are considering) :
(A ∩ B)’ = A – B.
But that is not true in general. If A is not the the universal set U ( and , in general, when one talks about two sets A and B, they are supposed to be different from the universal set) :
(A ∩ B)’ ≠ A – B.
As to your last question, the hypothesis of your problem does not allow you to determine whether A is or is not also included in B. Knowing that B is included in A does not rule out the possibility A to be also included in B, but it does not imply this either.
In case (a) neither A nor B is empty, (b) neither A nor B is the universal set and (b) the two sets are included one in the other , that is, in case they are equal , then :
(A ∩ B)’= (A∩ A)’ = A’ = complement of A = U – A.
A – B = A – A = ∅
So in that case : (A ∩ B)’ ≠ A – B