Summarising @DerekHolt's comments into an answer . . .
Finitely generated subgroups are closed under finite index in finitely generated groups, which is to say that if $H$ is such that $[G:H]<\infty $ and $G$ is finitely generated, then $H$ is finitely generated too. Thus the answer to 1 is "yes...
To summarise the comments into an answer . . .
Due to the Lemma, if $gA\cap g'A\neq \varnothing$, then $gA=g'A$.
Conversely, if $gA=g'A$, then $gA=gA\cap g'A(=g'A)$, and $gA$ is nonempty since $A$ is a group (and so contains $e$).
Promoting @ThomasAndrews' comment into an answer . . .
Yes, you're correct.
In general, if $G_1\le G_2$ and $\lvert G_2\rvert=p\lvert G_1\rvert$ for some prime $p$, then for any $G$ such that $G_1\le G\le G_2$, either $G=G_1$ or $G=G_2$.
To summarise the comments . . .
Your proof is fine, except that I recommend you consider $$\begin{align}
\varphi: H&\to gH \\
h&\mapsto gh,
\end{align}$$
then justify that $\varphi$ is a well-defined bijection; it's much easier than your $\phi$.
Also, Lagrange's Theorem only applies to finite ...
Yes, your proof is fine.
Using the arguments of the functions on the left is a little quaint; however, the notation is used in semigroup theory a lot, so I guess it's fine too.
Consider the cyclic group of order $3^3\cdot 5\cdot 7$.
In fact, for any $n\in\Bbb N$, there exists at least one groups, namely the cyclic group of order $n$.
Your proof is correct. Well done.
The phrasing at the end, though, is a little off; try a new sentence reading "Hence $\phi(G_1)$ is cyclic." It's just a minor suggestion.
