Note that $\mathbb{Q}$ is a divisible group: for every $x\in \mathbb{Q}$ and every $n\gt 0$, there exists $y\in\mathbb{Q}$ such that $ny=x$.
Therefore, any quotient of $\mathbb{Q}$ is also divisible. Thus, if $H$ is a subgroup of $\mathbb{Q}$, then $\mathbb{Q}/H$ is divisible.
But if $\mathbb{Q}/H$ is finite, then there exists $n\gt 0$ such that $n\left(\frac{a}{b}+H\right) = H$ for every $\frac{a}{b}\in\mathbb{Q}$; thus, if there is a nonzero element $\frac{r}{s}+H$ in $\mathbb{Q}/H$, can you find a $y+H\in\mathbb{Q}/H$ such that $n(y+H) = \frac{r}{s}+H$? What can we conclude from that?
To show every element is of finite order, let $x\neq 0$ be such that $x\in H$, and let $y\in\mathbb{Q}$. Show that there exists a $z\in\mathbb{Q}$ such that $\langle x,y\rangle = z$; conclude that there is a nonzero power of $y$ that is equal to a power of $x$ to deduce that $y+H$ has finite order in $\mathbb{Q}/H$.