If $A$ and $B$ are sets, then either $A \in B$ or $A\notin B$ Given that $A$ and $B$ are two sets, is the following proposition a tautology:
$A\in B \vee A\notin B$. 
I do not know any set theory beyond the naive one.
 A: This might depend on the form of logic you are using. It is certainly true if you use "standard" logic. And as has already been pointed out it just follows from basic logical axioms. 
Surprisingly though there are (fairly often) used logical systems where the statement $P\vee \neg P$ (called the law of excluded middle) is not provable. An example of this kind of logic is Intuitionistic logic.
I must admit that looking at ZFC axioms I don't see how to prove the statement in intuitionistic logic (then again I haven't really used it much). My feeling is that if it does follow it should in some way involve extensionality, but I can't figure out how.
A: Yes, though this has nothing to do with set theory really and is much more about propositional logic. If we replace "$A \in B$" with the propositional variable $\textrm{P}$, then your statement is simply an instance of 
$$\textrm{P} \lor \lnot \textrm{P}$$
which is of course always true.
A: Notice that 
$$(A\subset B)\vee (A\not\subset B)\iff (A\subset B)\vee(A\cap B^c\neq\emptyset)$$
which is always true.
