How to determine if this is true or false? $$\exists x \in X, (P(x) \to Q(x))\hspace{0.2cm} \iff (\exists x \in X, P(x))\to (\exists x \in X, Q(x))$$
$$\forall x \in X, (P(x) \to Q(x))\hspace{0.2cm} \iff (\forall x \in X, P(x))\to (\forall x \in X, Q(x))$$
Really don't know where to start 
 A: In general, to show that : $\varphi \leftrightarrow \psi$, we have to provide a proof in some proof system.
To show instead that  $\varphi \leftrightarrow \psi$ does not hold, we have to provide a suitable counter-example to one of the conditionals : $\varphi \rightarrow \psi$, $\psi \rightarrow \varphi$.

For :

$∀x(P(x) → Q(x)) \leftrightarrow (∀xP(x) → ∀xQ(x))$

we can provide a counter-example showing that :

$(∀xP(x) → ∀xQ(x)) \rightarrow ∀x(P(x) → Q(x))$

does not hold.
Consider an interpretation with domain the set $X = \{ 0,1 \}$ and consider as $P(x)$ the formula $(x = 0)$ and as $Q(x)$ the formula $(x > 0)$.
We have that :
$∀x(x = 0) → ∀x(x > 0)$
is true in $X$, because both the antecedent and the consequent are false.
Consider now :
$∀x[(x = 0) → (x > 0)]$;
choosing $0$ as value for $x$, we have that $(0 = 0) → (0 > 0)$ is false (antecedent true, while consequent false) and thus it is false that $(x = 0) → (x > 0)$ is true for any value of $x$.
Thus $∀x[(x = 0) → (x > 0)]$ is false in $X$.
In conclusion, we have found an interpretation such that $∀xP(x) → ∀xQ(x)$ is true and $∀x(P(x) → Q(x))$ is false, i.e. :


$∀xP(x) → ∀xQ(x) \nvDash ∀x(P(x) → Q(x))$,


and thus :

$\nvDash (∀xP(x) → ∀xQ(x)) \rightarrow ∀x(P(x) → Q(x))$.


The same interpretation can be used to show that :

$∃x(P(x) → Q(x)) \nvDash ∃xP(x) → ∃xQ(x)$.

Consider now as $P(x)$ the formula $(x > 0)$ and as $Q(x)$ the formula $(x < 0)$.
Now we have that $(x > 0) \rightarrow (x < 0)$ is true for $0$ as value of $x$, and thus $∃x[(x > 0) \rightarrow (x < 0)]$ is true in $X$.
But $∃x(x > 0)$ is true while $∃x(x < 0)$ is false, and thus $∃x(x > 0) \rightarrow   ∃x(x < 0)$ is false.
A: To try examples with these sentences, you have two predicates $P$ and $Q$.  Each member of $X$ can satisfy $P(x)$ or not and (separately) satisfy $Q(x)$ or not.  You can let $X$ have four members.  Each one will have one of the four possible truth values for $P$ and $Q$.  For example, one of them will have $P(x)$ true and $Q(x)$ false.  Now test each sentence for a tautology.  If one fails in this case you are done.  If a sentence passes, you need to think about whether deleting some element of $X$ will render it false.
A: Use Implication Equivalence and the Laws of Negation and Distribution of Quantifiers.
$$\begin{align}
 & \exists x \in X: (P(x) \to Q(x))
\\[1ex] \iff & \exists x\in X : (\neg P(x)\vee Q(x))
& \text{Implication Equivalence}
\\[1ex] \iff & (\exists x\in X: \neg P(x)) \vee (\exists x\in X:Q(x))
& \text{Distribution: Existential over Disjunction}
\\[1ex] \iff & \neg(\forall x \in X: P(x))\vee (\exists x \in X: Q(x))
& \text{Universal Negation}
\\[1ex] \iff & (\color{red}{\forall} x \in X:P(x))\to (\exists x\in X: Q(x))
& \text{Implication Equivalence}
\end{align}$$
So we must conclude that your given equivalence is not generally true.

For the other one, beware that the Distribution of Universal over Disjunction itself is not an equivalence.
