(i) $A= \mathbb{Q}/\mathbb{Z}$ the additive Group and I want to show first that $A_\text{tor}=\{a\in A:\text{ord}(a)<\infty \}=A$. $q\in \mathbb{Q} \Rightarrow q=\frac{a}{b}$ with $a,b \in \mathbb{Z}, b\neq 0$. Now : $q+\mathbb{Z} = \frac{a}{b} +\mathbb{Z}$. But $b\cdot (\frac{a}{b} +\mathbb{Z})=a+\mathbb{Z}=\mathbb{Z}$ what am I missing here? This seems far too easy.
(ii) If $A=(a_1,...,a_n)$ then from (i) it follows that the order of $a_i<\infty$, so there is a (if you want the smallest common multiple is enough) $k\in \mathbb{N}$ with $a_i^k=e, \, \forall a_i$. This follows from the fact that $A= \mathbb{Q}/\mathbb{Z}$ is abelian and ord$(a_1\cdot...\cdot a_n) | \text{lcd}(a_1,...,a_n) $ But beause the ord$(\frac{1}{n}+\mathbb{Z})=n$ for all $n\in \mathbb{N}, n\neq 0$, the order of elements of in $A= \mathbb{Q}/\mathbb{Z}$ cannot have an upper barrier like $\text{lcd}(a_1,...,a_n)$.
Is that ok?
Your proof is fine.
Let me just show a variant of (ii).
Let $H$ be the (finitely generated) subgroup od $A$ generated by $\bar a_1,...,\bar a_r$ with $a_i=\frac{m_i}{n_i}\in\Bbb Q$ and let $\cal P$ be the finite set of primes that are divisors of the $n_i$'s.
It is clear that if $\bar h\in H$ with $h=\frac mn\in\Bbb Q$ then the prime divisors of $n$ are in $\cal P$.
But there are infinitely many primes, so there exists $p\notin\cal P$ and $\overline{1/p}\notin H$.
Thus $H\neq A$, i.e. $A$ is not finitely generated.