There’s nothing wrong with the ‘vacuous truth’ part of the argument. It’s perfectly correct that if $X$ is any set, then $\left\{x\in X:x\in\bigcap\varnothing\right\}=X$. To see this, note that if $x\in X$, then $x\notin\bigcap\varnothing$ if and only if there is an $A\in\varnothing$ such that $x\notin A$, and since there is no $A\in\varnothing$ at all, this is not the case.
The problem with the argument is that nothing in $\mathsf{ZF}$ permits the formation of $\left\{x:x\in\bigcap\varnothing\right\}$: this an example of unrestricted comprehension, which is not permitted in $\mathsf{ZF}$. $\mathsf{ZF}$ permits only restricted comprehension, using a formula to pick elements from an already existing set, not from the universe at large.
For a) This would be the set of all numbers divisible by 2, or by 3, or by 4, ... etc. Since every number is divisible by itself, that means that this is the set of all numbers greater or equal to 2.
For b) This would be the set of all numbers divisible by 1, and by 2, and by 3, and by 4, ... etc. No number is like that, so this is the empty set.
To prove this you need to prove the following set equalities:
$\bigcup_{n=2}^{\infty}A_{n} = \{ n | n \ge 2\}$
$\bigcap_{n=1}^{\infty}A_{n} = \{ \}$
So for the first one, prove that any $n$ that is in $ \bigcup_{n=2}^{\infty}A_{n}$ is also in $\{ n | n \ge 2\}$, and vice versa.
The same for the second one ... except of course there is no such $n \in \{ \}$ ... so prove that there is no $n$ in $\bigcap_{n=1}^{\infty}A_{n}$ either. And that you could do by a proof by contradiction: suppose $n \in \bigcap_{n=1}^{\infty}A_{n}$. Then in particular $n \in A_{n+1}$, i.e. $n$ is divisible by $n+1$. But that is impossible, so we have a contradiction. So there is indeed no $n$ in $\bigcap_{n=1}^{\infty}A_{n}$.
Best Answer
Using $\mathbb{N} = \{1, 2, ...\}$.
For any $m \in \mathbb{N}$, $m \notin A_{m}$ since $A_{m} = \{m + 1, (m + 1)(2), ... \}$. Hence $m$ is not in the intersection $\bigcap_{n = 1}^\infty A_n$. Thus the intersection does not contain any $m \in \mathbb{N}$.