Warning: Not of concern to the OP.
I will differ from ultrafilter here, and note: The answers to all three questions are yes (taking “contain” in the ∈ sense).
First, note that X and Y are both sets (I will use this fact implicitly in the following).
The first question is the trickiest: Y ∈ Y ⇔ Y ⊆ X ⇔ each element of Y is an element of X ⇔ each subset of X is a set. But all subsets are trivially sets, so this is true.
(Note, also, that trivially, Y ⊆ Y)
The second question: X ∈ Y ⇔ X ⊆ X, which is trivially true.
(Note, however, that X ⊆ Y ⇔ each element of X is an element of Y ⇔ each set is a subset of X ⇔ each set has only sets as elements. This depends on the sense of the word “set” one uses. If one allows non-sets (ur-elements) in sets, this is false. Otherwise, this is true.)
The third question: Y ∈ X ⇔ Y is a set, which is even more trivially true.
(Note, also, that Y ⊆ X ⇔ each element of Y is an element of X ⇔ each subset of X is a set, which is trivially true)
Now, one may finally object “But there’s no such thing as the set of all sets”; this is true on some set theories, but not in keeping with the spirit of the question. We can perfectly well make a consistent theory in which there are objects called sets, a binary relation ∈ on them, a set is uniquely determined by what is ∈ it, and there are universal and power-sets. Most trivially, there is the model with just a single set which contains itself. Less trivially, we may take something like New Foundations (with or without urelements). The ZF-style sense of “set” as meaning specifically “well-founded extensional tree, bounded in branching width (equivalently, in rank) by some inaccessible cardinal” is just one among many.