Proof using existential quantifier Prove: 
$$\begin{align} \exists x ~:~ \bigg(p(x) &\rightarrow q(x)\bigg) \
How do I go about proving this? Can I distribute the existential quantifier in the first term? 
 A: Two relevant laws (whether they are laws of inference or not depends on your logic) would be:
$$A \rightarrow B \equiv \lnot A \lor B \tag{1}$$
$$\exists y ~:~ m(y) \lor n(y) \equiv \exists y ~:~ m(y) \lor \exists y ~:~  n(y)\tag{2}$$
Property (1) may as well be the definition of $\rightarrow$.  Property (2) is the result of $\exists$ being a quantified $\lor$.
To your question:

Can I distribute the existential quantifier in the first term?

Let's apply properties (1) and (2) and see:
$$\exists x ~:~ p(x) \rightarrow q(x)$$
$$\exists x ~:~ \lnot p(x) \lor q(x)$$
$$\left(\exists x ~:~ \lnot p(x)\right) \lor \left(\exists x ~:~ q(x)\right)$$
$$\lnot \left(\exists x ~:~ \lnot p(x)\right) \rightarrow \left(\exists x ~:~ q(x)\right)$$ 
$$\left(\forall x ~:~ p(x)\right) \rightarrow \left(\exists x ~:~ q(x)\right)$$ 
So you see what happens when you try to distribute the $\exists$ across a $\rightarrow$.

How do I go about proving this?

Simplify your expression.
$$\begin{align} \exists x ~:~ \bigg(p(x) &\rightarrow q(x)\bigg) \\
 &\rightarrow\\
 \bigg(\exists x ~:~ p(x) &\rightarrow \exists x ~:~ q(x)\bigg)\end{align}$$
Convert $\rightarrow$ to $\lor$:
$$\begin{align} \lnot(\exists x ~:~ \lnot p(x) &\lor q(x)) \\
 &\lor\\
 \bigg(\lnot \exists x ~:~ p(x) \bigg) &\lor \exists x ~:~ q(x)\end{align}$$
Carry through the $\lnot$ to the leaf terms:
$$\begin{align} (&\forall x ~:~ p(x) \land \lnot q(x)) \\
 &\lor \bigg(\forall x ~:~ \lnot p(x) \bigg) \\
 & \lor \exists x ~:~ q(x)\end{align}$$
Now you can probably find a counter example.  For a counter example, all 3 of the above or cases should be false.  The third case:
$$\exists x ~:~ q(x)$$
implies that a counter example must have $q(x) = \text{false}$.  So to find a counter example for:
$$\begin{align} (&\forall x ~:~ p(x)) \\
 &\lor \bigg(\forall x ~:~ \lnot p(x) \bigg) \\
 \end{align}$$
So your counter examples will be those situations where $p(x)$ is sometimes true and sometimes false, and when $q(x)$ is false.  In all other cases the statement is true.
