$o(g^{k})=\frac{o(g)}{\gcd(o(g),k)}$ order of a power of a group element.

By definition, $\operatorname{ord}(g^k)$ is the least $l\geq 1$ such that $(g^k)^l=1$; but since $\operatorname{ord}(g)=n$, $1=(g^k)^l=g^{kl}$ if and only if $n|kl$. This proves a).

As you said, b) is simply a consequence of a) : indeed a) tells us that $k\operatorname{ord}(g^k)$ is the least multiple of $k$ that is also a multiple of $n$, and thus the least common multiple of $k$ and $n$; thus $$\operatorname{ord}(g^k)=\frac{LCM(n,k)}{k}=\frac{nk}{GCD(n,k)k}=\frac{n}{GCD(n,k)},$$ and thus $\operatorname{ord}(g^k)=n\Leftrightarrow GCD(n,k)=1$.

Hint $\ g^{\large kx}=1 \!\iff\! n\mid kx\!\iff\! n\mid kx,nx\!\iff\! n\mid (kx,nx)\!=\! (k,n)x\!\iff\! n/(k,n)\mid x\ $

Since $\,\ell = {\rm ord}\,g^{\large k}\,$ is the least such $\,x,\,$ clearly $\ \ell = n/(k,n),\ $ so $\ \ell = n\iff (k,n)= 1.\,$