What restrictions lay on truth sets of predicates A(x) and B(x) if the given predicate is true Suppose we are given 2 predicates
$A(x)$ and $B(x)$ with domain $M$.
Suppose next we are given the following predicate
$$\neg (A(x) \land B(x)) \land (\forall x(A(x) \rightarrow B(x)))$$
which we know is true, so
$$\neg (A(x) \land B(x)) \land (\forall x(A(x) \rightarrow B(x))) = 1$$
The question is how does it restrict the truth sets of $A(x)$ and $B(x)?$
It is obvious that we have
$$\neg (A(x) \land B(x)) = 1 \\
A(x) \land B(x) = 0\\
A(x) = 0 \lor B(x) = 0$$
So from that we get that either truth set for $A(x)$ is $E_A \neq M$ or truth set for $B(x)$ is $E_B \neq M$.
But knowing that
$$\forall x(A(x) \rightarrow B(x)) = 1\\
\forall x(\neg A(x) \lor B(x)) = 1$$
I have no idea how to link it to useful information on truth sets of $A(x)$ and $B(x)$, any suggestions?
You are getting hung up on truth sets when there is only one variable in the problem.  Focus on one element $x$ of $M$ and ask whether $A(x)$ and $B(x)$ can be true because all the elements are equivalent.  You have found that both $A(x)$ and $B(x)$ cannot both be true but $A(x) \implies B(x)$.  You should be able to derive that $A(x)$ is false and $B(x)$ can be anything.  Check that in the original axiom and it works.
1
$$\neg (A(x) \land B(x)) \land (\forall x(A(x) \rightarrow B(x)))$$
2
$$\neg (A(x) \land B(x)) \equiv (\neg A(x)) \lor (\neg B(x))$$
$$(\forall x(A(x) \rightarrow B(x))) \equiv (\neg A(x)) \lor B(x)$$
3
$$(\neg A(x) \lor \neg B(x)) \land (\neg A(x) \lor B(x)) \equiv (\neg A(x))$$