$a)$ is obvious. $b)$ Note that all elements of $H$ are of finite order, thus they are all of the form $e^{\frac{2\pi ik}{n}}$, where $(k, n) = 1$, this automatically implies that $e^{\frac{2\pi i}{n}} \in H$. Since $H$ is infinite, denomenators in elements of the form $e^{\frac{2\pi i}{n}}$ are unbounded. If we prove that these denomenators are divisible by only one prime number $p$, then $H \cong G_{p}$. Try it yourself, i'll post that part later if noone succeeds.

What is $(\mathbb{C}^*, \cdot).?$

Dey wrote: What is $(\mathbb{C}^*, \cdot).?$ The complex numbers but with a multiplicative group structure, excluding 0.