Checking validity of formulas in first-order logic
First of all, from this post

Heuristic: To try to determine whether ϕ is valid: If the answer's not
obvious, look for an interpretation where ϕ is false.

So try to proof by negation. Consider $1:$
Let
\begin{gather*} 
\phi\equiv(\exists xP(x)\to Q )\to(\forall x(P(x)\to Q)).
\end{gather*}
Now negate $\phi$ then we have:
\begin{gather*} 
\neg\phi\equiv \neg(\neg(\exists xP(x)\to Q )\vee(\forall x(P(x)\to Q)))\equiv (\exists xP(x)\to Q )\wedge \neg(\forall x(P(x)\to Q))\equiv (\forall xP(x)\vee Q )\wedge (\exists x(P(x) \wedge \neg Q))
\end{gather*}
so after elimination of quantifiers:
\begin{gather*} 
 \neg P(x)\vee Q 
\end{gather*}
\begin{gather*} 
P(C_1)
\end{gather*}
\begin{gather*} 
\neg Q 
\end{gather*}
oblivious that if we use resolution, and resolve each clause we achieve $\bot$.
As a result we can say $1$ is valid.
$2$ is not valid, you can use above method or if we set
\begin{gather*} 
Q\equiv \top
\end{gather*}
\begin{gather*} 
 P\equiv R\equiv \bot
\end{gather*}
as a result $2$ is $\bot$.
Repeat idea that used in $1$ for $3$:(as you know from logic, $A\leftrightarrow B\equiv (A\to B) \wedge  (B\to A)$) so we have:
\begin{gather*} 
 \neg(((\exists x(P(x)\to Q(x)) \to (\forall xP(x)\to \exists xP(x)))
\end{gather*}
\begin{gather*} 
\wedge
\end{gather*}
\begin{gather*} 
(\forall xP(x)\to \exists xP(x)) \to ((\exists x(P(x)\to Q(x))  ))
\end{gather*}
After applying  negation:
\begin{gather*} 
\equiv
\end{gather*}
\begin{gather*} 
 (((\exists x(\neg P(x)\vee Q(x)) \wedge (\forall xP(x)\wedge \forall  x\neg P(x)))
\end{gather*}
\begin{gather*} 
\vee
\end{gather*}
\begin{gather*} 
(\exists x\neg P(x)\vee \exists xP(x)) \wedge  ((\forall x(P(x)\wedge \neg Q(x))  ))
\end{gather*}
Eliminate quantifiers:
\begin{gather*} 
 (\neg P(C_1)\vee Q(C_1)) \wedge (P(x)\wedge\neg P(x))
\end{gather*}
\begin{gather*} 
\vee
\end{gather*}
\begin{gather*} 
(\neg P(C_2)\vee P(C_3)) \wedge  (P(x)\wedge \neg Q(x)  ))
\end{gather*}
As you know
\begin{gather*} 
 (\neg P(C_1)\vee Q(C_1)) \wedge (P(x)\wedge\neg P(x))\equiv \bot
\end{gather*}
So we have:
\begin{gather*} 
 \bot
\end{gather*}
\begin{gather*} 
\vee
\end{gather*}
\begin{gather*} 
(\neg P(C_2)\vee P(C_3)) \wedge  (P(x)\wedge \neg Q(x)  ))
\end{gather*}
\begin{gather*} 
\equiv
\end{gather*}
\begin{gather*} 
(\neg P(C_2)\vee P(C_3)) \wedge  (P(x)\wedge \neg Q(x)  ))
\end{gather*}
As a result above formula can't equal to $\top$ because we can't resolve some clauses to get $\bot$. So $3$ isn't valid.
$4$ remain as exercise, if you follow described approach for other options, you can see $4$ is not valid.