$A1: \exists x [A(x) \wedge B(x)] \rightarrow \exists x [A(x)] \wedge \exists x [B(x)] $
It is valid
You can try to create $T \rightarrow F$ result but you cannot because of there exist x which is true for both A and B then that same x on RHS will also hold the property A and B.
$A2: \exists x \forall y Q(x,y) \rightarrow \forall y \exists x Q(x,y)$
This is also valid.
LHS of A2 says that $\text{"There exist x for all y for which Q(x,y) holds"}$
Lets assume $Q(x,y)$ means x is a friend of y,
Then LHS of A2 says that there is someone x who is friends with everyone.
and RHS of A2 says that $\text{"For every y there is some x for which Q(x,y) holds"}$
which can be interpreted as for every person y there is someone (x) such that x and y are friends.
You can try to create $T \rightarrow F$ result but you cannot because if LHS is true that means there is someone who is friends with everyone and then RHS cannot be false. for RHS making false there must be someone who doesn't have any friend but LHS says that there is someone who is friends with everyone.