I recently read a proof of why the null set is contained in every set, a concept which heretofore had confused me.
In formal logic, if the antecedent of an implication is false, then the implication as a whole is true. This is because the truth value of the consequent is independent of the truth value of the antecedent (though the converse of this statement is false).
With this in mind, consider the following implication:
$x$ $\in$ $\emptyset$ $\implies$ $x$ $\in$ $S$,
where $S$ is an arbitrary set. By the definition of the empty set, the antecedent is always false. Thus the implication itself is always true.
This is why your reasoning behind (1) is false. The reason "false" is the correct answer is because $X$ $=$ $\emptyset$; in other words, $X$ contains nothing. The null set is something, and so $X$ cannot contain it.
(2) is true because $Y$ clearly contains $\emptyset$.
(3) is technically true, for the argument given above. But suppose $Z$ is some exceptional, nonexistent set that does not contain the null set, which seems to be the assumption of whoever wrote this problem. The message being conveyed is that a set containing the null set is not the same as the null set itself, which is why it is false, even though such a set does not, and cannot, actually exist.
(4) is true, but vacuously so. This is because $X$ $\subseteq$ $Y$ can be rewritten as $x$ $\in$ $X$ $\implies$ $x$ $\in$ $Y$. As shown above, this implication is true because the antecedent is always false.
(5) is actually false. The notation here is a bit subtle. If $A$ and $B$ are arbitrary sets, then, if $A$ $\subseteq$ $B$, $x$ $\in$ $A$ $\iff$ $x$ $\in$ $B$, for all possible $x$. However, if $A$ $\subset$ $B$, $A$ is known as a "proper subset" of $B$, and implies that there exists at least one element $x$ such that $x$ $\in$ $B$ $\land$ $x$ $\notin$ $A$; in other words, all elements of $A$ are contained in $B$, though the converse of this statement is false. It may be the case that whoever wrote these problems uses $\subseteq$ to mean "proper subset of", in which case (5) is true.
(6) is true for the same reason (4) is false.
(7) is true for the same reason (6) is true.
It is important to make the distinction between $X$ $\subset$ $Y$, which means all elements of $X$ are contained in $Y$, and $X$ $\in$ $Y$, which means the set $X$ itself, though not necessarily the individual elements of $X$, is contained in $Y$.