Can you help me verify if I answered this question correctly?
Consider
$[(\forall x)(P(x)) \land (\exists x)(\lnot Q(x))] \implies \{(\forall x)(P(x)) \iff [\lnot(\forall x)(R(x)) \lor (\exists x)(S(x))]\}$
If you know that this proposition is false, then determine the truth values of the following:
$(\forall x)(P(x))$
$(\forall x)(Q(x))$
$(\exists x)(\lnot R(x))$
$(\exists x)(S(x))$
Since we know that this implication is false, it must be of the form $V_0\implies F_0$
Since the rightmost proposition is false, we should negate it. That is to say,
$\lnot \{(\forall x)(P(x)) \iff [\lnot(\forall x)(R(x)) \lor (\exists x)(S(x))]\}$
The negation of $P \iff Q$ is $(P \land \lnot Q) \lor (\lnot P \land Q)$, so the above becomes this:
$\{(\forall x)(P(x)) \land \lnot[\lnot(\forall x)(R(x)) \lor (\exists x)(S(x))]\} \lor \{\lnot(\forall x)(P(x)) \land [\lnot(\forall x)(R(x)) \lor (\exists x)(S(x))]\}$
$\{(\forall x)(P(x)) \land [(\forall x)(R(x)) \land \lnot (\exists x)(S(x))]\} \lor \{\lnot(\forall x)(P(x)) \land [\lnot(\forall x)(R(x)) \lor (\exists x)(S(x))]\}$
$\{(\forall x)(P(x)) \land [(\forall x)(R(x)) \land (\forall x)(\lnot S(x))]\} \lor \{(\exists x)(\lnot P(x)) \land [(\exists x)(\lnot R(x)) \lor (\exists x)(S(x))]\}$
Now, if you look at the rightmost part of this disjunction, you will see that we got a conjunction. One of the propositions in this conjunction is $(\exists x)(\lnot P(x))$, but that happens to be false because we know that $(\forall x)(P(x))$. So the whole conjunction is false, and by disjunctive syllogism we end up with only the leftmost part of the disjunction:
$\{(\forall x)(P(x)) \land [(\forall x)(R(x)) \land (\forall x)(\lnot S(x))]\}$
At the end, we got all of these facts:
$(\forall x)(P(x)) \land (\exists x)(\lnot Q(x)) \land (\forall x)(R(x)) \land (\forall x)(\lnot S(x))$
So the answers to the question would be:
$(\forall x)(P(x)) \equiv V_0$
$(\forall x)(Q(x)) \equiv F_0$
$(\exists x)(\lnot R(x)) \equiv F_0$
$(\exists x)(S(x)) \equiv F_0$
You can use Wolfram Alpha to generate a truth table, provided you know the correct syntax. Here is the table for your schema.
We identify $$ p = (\forall x) P(x)$$ $$q = (\forall x) Q(x)$$ $$ r = (\forall x) R(x) $$ $$s = (\exists x) S(x)$$
The table shows that there is only 1 assignment that makes the statement false, namely $$ p \equiv T, \ q \equiv F, \ r\equiv T, \ s \equiv F$$ which gives $$(\forall x) P(x) \equiv T$$ $$(\forall x) Q(x) \equiv F$$ $$\neg (\forall x) R(x) \equiv (\exists x)(\neg R(x)) \equiv F$$ $$(\exists x) S(x) \equiv F$$ the same as you surmised originally.