For (1), suppose that $H$ is a subgroup of $G$ of order $d$. Let $y\in H$. Since $G$ is cyclic with generator $x$, you know that $y=x^k$ for some integer $k$ such that $0\le k<n$. Show that $k$ is a multiple of $m$ and conclude that $y$ is in the subgroup generated by $x^m$. Since $y$ was an arbitrary element of $H$, it follows that $H$ is contained in the subgroup generated by $x^m$, and since these sets both have $d$ members, they must be equal. There is a standard trick to proving that $k$ is a multiple of $m$: use the division algorithm to write $k=mq+r$, where $q$ and $r$ are integers and $0\le r<m$, and show that $r$ must be $0$. To do that you’ll want to use the fact that since $H$ has order $d$, $y^d=1_G$.
For (2), you know that $b$ is a generator of $G$ if and only if the order of $b$ is $n$. You know that $y^n=1_G$ for every $y\in G$, so the order of $b$ is $n$ if and only if no smaller positive power of $b$ is the identity. Thus, you want to show two things:
$\qquad\qquad\qquad(a)\quad$ if $\gcd(n,k)=1$, and $0<m<n$, then $b^m\ne 1_G$,
$\qquad\qquad\qquad(b)\quad$ if $\gcd(n,k)>1$, then the order of $b$ is less than $n$.
The first implies that whenever $\gcd(n,k)=1$, $b$ is a generator of $G$; the second implies that whenever $\gcd(n,k)>1$, $b$ is not a generator of $G$.
For $(b)$, think about what you did in problem (1). For $(a)$, you need to show that $km$ is not a multiple of $n$.