1) Define a homomorphism $\varphi$: $\mathbb{Z}\rightarrow G$ such that $\varphi(n)=x^m$. Then $\varphi$ is surjective. Consider the kernel of $\varphi$, $N=\ker\varphi$. Since $G$ is of order $n$, $N=n\mathbb{Z}.$ By the isomorphism theorem,
$$G\cong\mathbb{Z}/n\mathbb{Z}.$$
Applying the isomorphism theorem again, the subgroups of $G$ corresponds to the subgroups $t\mathbb{Z}$ in $\mathbb{Z}$ such that $t|n,~s>0.$ The image of these subgroups in $G$ are $\langle x^t\rangle$ with order $n/t= d$.
2) I will show a more general case. $\langle x^k\rangle$ is a subgroup of $G$. As 1) shows, $\langle x\rangle$ has subgroups of the form $\langle x^d\rangle$ with $d|n$. Since $\langle x^k\rangle=\langle x^d\rangle$, we have $$k=ad+bn,$$ $$d=uk+vn,$$ where $a,b,u,v$ are integers. Then we obtain that $d|k$ and $d=(n,k)$. Applying 1), we have the order of $\langle x^k\rangle$ is $n/d$, so is the order of $x^d$. Your question is the case for $d=1$.