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Let $H,K$ be the normal subgroups of orders $3,5$. Their intersection is a subgroup of both, so Lagrange's theorem makes its order divide both $3$ and $5$; hence $H\cap K=\{1\}$. For $h\in H$ and $k\in K$, normality shows that $hk{h}^{-1}{k}^{-1}$ belongs to both $H$ and $K$. It must be the identity, so $hk=kh$. This proves the needed instance of normal subgroups of coprime order commute.

Choose nonidentity $h\in H$ and $k\in K$. Their orders are $3$ and $5$, since these orders divide the respective prime subgroup orders. If $(hk{)}^m=1$, commutativity gives $h^m=k^{-m}\in H\cap K$, so $3\mid m$ and $5\mid m$. Conversely $(hk{)}^{15}=1$. Thus the product of commuting elements of coprime order gives an element $hk$ of order $15$.

For the counterexample, take the dihedral group of symmetries of a regular pentagon:Its ten elements are $r^j$ and $s{r}^j$, $0\le j<5$. Nonidentity rotations have order $5$, and $(s{r}^j{)}^2=r^{-j}{r}^j=1$, so all reflections have order $2$. There is no element of order $10$.