I have the following:
$∀x(P(x) \land Q(x)) ≡ ∀xP(x) \land ∀xQ(x)$
$∀x(P(x) \lor Q(x)) ≡ ∀xP(x) \lor ∀xQ(x)$
How do I disprove or prove that they're equivalent? My attempt the first one is that if I called $P(x) =$ "x is even" and $Q(x) = $ "x is odd", how can the LHS be true? A number could not be even and odd at the same time so it would evaluate to false. How does this apply for the RHS and the second part? I'm not sure how to interpret these.
Since FOL is sound & complete, it doesn't matter (unless this is a homework task) whether you're going to prove this using FOL's semantics (truth trees etc.) or syntactic inference rules (of some appropriate axiomatization).
$\forall x(P(x) \land Q(x)) \equiv \forall xP(x) \land \forall xQ(x)$ is actaully two things:
And these two are really:
To prove the first, assume there is a model $\mathcal{M}$ in which it doesn't hold. Then (for this model), by definition of $\rightarrow$, $\forall x(P(x) \land Q(x))$ but $\neg(\forall xP(x) \land \forall xQ(x))$. From $\forall x(P(x) \land Q(x))$ by definition of $\forall$: for all $a \in \mathcal{M}$, $P(a) \land Q(a)$, and now by definition of $\land$, for all $a \in \mathcal{M}$, $P(a)$, that is, by definition of $\forall$, $\forall xP(x)$. Do the same thing with $Q$ to get $\forall xQ(x)$. Now we have by definition of $\wedge$: $\forall xP(x) \land \forall xQ(x)$. Earlier we got $\neg(\forall xP(x) \land \forall xQ(x))$, contradicting the new result. So there is no such model $\mathcal{M}$, so (1) holds.
The (2) is almost the same, try it yourself.
Your second statement doesn't hold, it's enough to construct a model in which it doesn't hold. If we interpret $P$ as "is a woman", and $Q$ as "is a man", and we let the females and the males of Earth populate our $\mathcal{M}$, the left side holds, but the right side doesn't.