Computing $\pi_{et}(X,x)^{\operatorname{ab}}$

Question

Let $X$ a smooth projective algebraic curve of genus $g$ over $k$. ($k$ is an algebraically closed field of characteristic $0$).

I want to compute $\pi_{et}(X,x)^{\operatorname{ab}}$.

I'm trying to understand the proof presented here at page 28. http://math.univ-lille1.fr/~borne/Recherche/pisa.pdf

Here are the steps of the proof:

1. It is enough to understand $\operatorname{Hom}(\pi_{et}(X,x),A)$ where $A$ is a finite abelian group.

2. $\operatorname{Hom}(\pi_{et}(X,x),A) = H^1(X,A)$

3. and we can stick to the case $A = \mu_n$.

4. Using Kummer theory $H^1(X, \mu_n) = (\mathbb{Z}/n)^{2g}$

5. So, we get $\pi_{et}(X,x)^{\operatorname{ab}}= (\hat{\mathbb{Z}})^{2g}$

I understand all the steps except the first and last step. Can you someone why it is enough to understand $\operatorname{Hom}(\pi_{et},\mu_n)$? And we exactly in last step we get $\pi_{et}(X,x)^{\operatorname{ab}}= (\hat{\mathbb{Z}})^{2g}$?

It seems they are claiming $$\pi_{et}(X,x)^{\operatorname{ab}} = \underset{n}{\varprojlim} \operatorname{Hom}(\pi_{et}(X,x),\mu_n) $$?

Why is the last statement true? Does it hold for any profinite group?