Show that $x+\mathbb{Z}$ of $\mathbb{R}/\mathbb{Z}$ has finite order iff $x\in \mathbb{Q}$.
I don't really see how this follows. Say, $|x+\mathbb{Z}| = n$, first of all, how can it have finite order - $\mathbb{Z}$ is infinite. I feel like I'm not understanding some very important concept in quotient groups.
An element of an infinite group can have finite order; if the group is written additively, this happens if and only if $ng=0$ for some $n>0$.
In your case, if the order of $x+\mathbb{Z}$ is $n$, then $$ n(x+\mathbb{Z})=nx+\mathbb{Z}=0+\mathbb{Z} $$ which means $nx\in\mathbb{Z}$, hence $x\in\mathbb{Q}$.
Conversely, suppose $x=\frac{m}{n}\in\mathbb{Q}$. Can you go on and prove the element $x+\mathbb{Z}$ has finite order?