Showing that $\operatorname {Br}(\Bbb F_q)=0$ I want to prove that $\operatorname {Br}(\Bbb F_q)=0$ using the cohomological description of the Brauer group.
We have:
$\operatorname {Br}(\Bbb F_q)=H^2(\operatorname {Gal}(\overline {\Bbb F_q}/\Bbb F_q), \overline {\Bbb F_q}^*)$.
Since $\operatorname {Gal}(\overline {\Bbb F_q}/\Bbb F_q)$ is just $\widehat {\Bbb Z}$, this is just 
$H^2(\widehat{\Bbb Z}, \overline {\Bbb F_q}^*) = \operatorname{Ext}_{\Bbb Z[\widehat{\Bbb Z}]}^2(\Bbb Z, \overline {\Bbb F_q}^*)$.
So I just need to find a free $\Bbb Z[\widehat{\Bbb Z}]$-resolution of $\Bbb Z$:
$$...\to \Bbb Z[\widehat{\Bbb Z}] \to \Bbb Z\to 0$$
But already the kernel of the first map is complicated. Any suggestions?
EDIT: as suggested by @Mariano, I'll try another approach.
Assume $\operatorname {Br}(\Bbb F_q)\neq 0$ and let $D$ be division algebra over $\Bbb F_q$. Then $\dim_{\Bbb F_q} D = n^2$ for some integer $n$. Let $l=q^{n^2}$. Then it suffices to show that $\operatorname{Br}(\Bbb F_l / \Bbb F_q) =0$. $\operatorname {Gal}({\Bbb F_l}/\Bbb F_q)=\Bbb Z/n^2\Bbb Z$, so I need to compute $\operatorname{Ext}_{\Bbb Z[\Bbb Z/n^2\Bbb Z]}^2(\Bbb Z, \overline {\Bbb F_l^*})$

That cohomology group of the Galois group is not normal cohomology but continuous cohomology with respect to the Krull topology on the Galois group. In particular, your approach does not have many chances.
Now, the definition of both Krull topology and of continuous cohomology are made so that you can reduce everything to computing cohomology of finite quotients of the Galois group, which are cyclic. You should consult a textbook on Galois cohomology.
