My attempt for the first: (I would like to get it verified because I didn't use property of a cyclic group) $|G|<\infty$ (since for otherwise $(a^2),(a^3)$ are distinct improper nontrivial subgroups) can't have more than one prime divisor since for otherwise $G$ would contain more than one element of those order (and hence subgroups which must be nontrivial proper).
Since $|G|\ne 1,~|G|=p^k$ for some prime $p$ and for some $k\ge1.$
The existence of $H$ ensures $k>1.$
However if $k>2$ by Sylow theorem $G$ would contain at least two proper subgroups of orders $p,p^2.$
So $k=2.~\Box$
My attempt for the second: Since the argument for the first didn't use the property of a cyclic group, $|G|=p^2$ for some prime $p.$ Consequently $G$ is abelian and hence the isomorphic classed are $$\mathbb Z_{p^2}\\\mathbb Z_p\times\mathbb Z_p$$Since $\mathbb Z_p\times\mathbb Z_p$ has at least two proper subgroup viz. $\langle(1,0)\rangle,\langle(0,1)\rangle$ we have $G\simeq\mathbb Z_{p^2},$ cyclic $\Box.$
Please tell me if I'm correct!!
Best Answer
For the first one you might use the fact that the number of subgroups of a cyclic group of order $n$ equals the number of divisors $d(n)$ of $n$. For a prime $p$, $d(p^k)=k+1$, and hence $k=2$.
For the second statement: pick any $g \in G$, with $g \neq 1$. Then $<g>$ is non-trivial, hence equal to $G$ or a proper subgroup. In the latter case pick $h \in G\backslash<g>$. Then we must have $<h>=G$. So $G$ is cyclic either way.