$|a|=m,\,\gcd(m,n)=1 \implies a$ is an $n$'th power

HINT: Let $H$ be the subgroup of $G$ generated by $a$ (so $H=\{a^1, a^2, . . . , a^m=e\}$). Consider the map $f: H\rightarrow H$ given by $f(x)=x^n$. Can you show that $f$ is injective? Do you see why the injectivity of $f$ solves your problem?

SUBHINT: Suppose $f$ weren't injective. Then $a^{xn}=a^{yn}$ for some distinct $x, y$ in $\{1, . . . , m\}$. Suppose WLOG that $x<y$. Then $a^{(y-x)n}=e$ (why?). If $n$ is coprime to $m$, do you see why this yields a contradiction?

It will be useful to prove the following:

If $a$ has order $n$, then $a^k=e$ iff $n\vert k$.

Hint: Euclid's algorithm - If $n$ and $m$ are coprime we can find $p,q \in \mathbb{Z}$ such that $pn + qm = 1.$ Then, $1=a^{qm} = a^{1-pn}.....$ can you continue from here?