Proof attempt:
(a) $g^ng^m=g^{n+m}$
(b) $(g^n)^{-1}=g^{-n}$
Proof(informal rough draft).
(a) Since $g\in G$, we can rewrite $g^n=gg...g$ for n-factors of $g$ and $g^m=gg...g$ for m-factors of $g$, since rules of exponents still hold for groups(given definition in the book). If $G$ is under a group under addition, we can add the exponents to obtain $g^ng^m=g^{n+m}$.
Have not yet began (b).
Thoughts? This is an intro to group theory/proof class so instructor said it should be simple yet very direct.
Your proof of part (a) is, as mentioned in the comments, not a proof; you are assuming that which you wish to show. Furthermore, writing "If $G$ is under a group [sic] under addition, we can add the exponents to obtain $g^n g^m = g^{n+m}$ is not just assuming that which you want to show, but also is misleading: this identity holds in any group, and the question asks you to show that it indeed holds for any group, not just a group with addition as its operation.
As this looks like homework, I'll give a hint for each question.
For part (a), you say that the definition of $g^n$ that you were given is as the product of $n$ copies of $g$. What happens if you multiply $n$ copies of $g$, and then multiply this by $m$ copies of $g$? How do you write this product in your group, before and after you carry out the multiplication?
For part (b), as stated in the comments, it suffices to show that $g^n g^{-n}$ is the identity element. As (presumably, analogous to the definition in (a)) the definition of $g^{-n}$ given is the product of $n$ copies of $g^{-1}$, what would you get if you first multiply $n$ copies of $g$, followed by $n$ copies of $g^{-1}$? How would you write either product in $G$?