For #1, let $\displaystyle N_1,N_2$ be normal subgroups of orders $\displaystyle 3\mbox{ and }5$ respectively. Using the identity $\displaystyle |N_1N_2||N_1\cap N_2| = |N_1||N_2|$ and the fact that $\displaystyle N_1\cap N_2 = \{ e\}$ we find that $\displaystyle N_1N_2$ is a subgroup of order $\displaystyle 15$. Let $\displaystyle N_1N_2 = H$ then $\displaystyle N_1,N_2\triangleleft H$ with $\displaystyle N_1\cap N_2 = \{ e\}$ this means $\displaystyle H\simeq N_1\times N_2 \simeq \mathbb{Z}_3\times \mathbb{Z}_5\simeq \mathbb{Z}_15$. Therefore this group is cyclic of order $\displaystyle 15$.
Note, $\displaystyle D_5$ has ten elements but it is not cyclic.
The general result says that the subgroups of $\displaystyle \mathbb{Z}_n$ (the cyclic subgroup of order $\displaystyle n$) are precisely $\displaystyle k\mathbb{Z}_n = \left< [k]_n\right>$ where $\displaystyle k$ is a positive divisor of $\displaystyle n$. And furthermore, $\displaystyle k\mathbb{Z}_n$ is a subgroup of $\displaystyle m\mathbb{Z}_n$ if and only if $\displaystyle m|k$.
For example, the complete list of subgroups of $\displaystyle \mathbb{Z}_{24}$ are given below.