Show that the formulas have even number of paranthesis 
Using structural induction,
Show: All propositional logic expressions contain an even number of parentheses. (0 is considered an even number)

Base Case: Let $p$ be a propositional letter, then we have $p$ has 0 parantheses, thus even.
Inductive: We show for complex-er statements.
(1) Take: $(\neg p) \space$ then How do I use the hypothesis?
I don't see a way of formally saying $(\neg p)$ must have even # of parantheses, so what is the inductive way?

You do structural induction on this one, meaning that you prove that:
Base case: it is true for all atomic propositions (you did this, good!)
Step:  You assume (inductive hypothesis) that the claim is true for $\varphi$ and $\psi$, and show that the claim will be true for $\neg \varphi$, $(\varphi \land \psi$), $(\varphi \lor \psi)$, $(\varphi \rightarrow \psi)$, and $(\varphi \leftrightarrow \psi)$
