This is not going to be a reference, but I think the following is a great exercise (using étale cohomology) to prove the fundamental exact sequence for Brauer groups of global fields.
Theorem: Let $X/\mathbf{F}_q$ be a smooth proper geometrically connected curve and $K$ its function field. Then there is an exact sequence $$ 0 \to \text{Br}(K) \to \bigoplus_{\nu} \text{Br}(K_\nu) \to \mathbf{Q}/\mathbf{Z} \to 0.$$
To prove the theorem, consider the divisor exact sequence $$ 0 \to \mathbf{G}_{m,X} \to j_\ast \mathbf{G}_{m, \eta} \to \bigoplus_{x \in |X|} (i_x)_\ast \mathbf{Z} \to 0.$$ Taking cohomology, we get $$ H^2(X,\mathbf{G}_m) \to H^2(X, j_\ast \mathbf{G}_m) \to \oplus_{x \in |X|} H^2(X, (i_x)_\ast \mathbf{Z})\to H^3(X, \mathbf{G}_m) \to H^3(X, j_\ast \mathbf{G}_m).$$
We must now show that: $$\begin{eqnarray} H^2(X,\mathbf{G}_m) &=& 0 \\ H^2(X, j_\ast\mathbf{G}_m) &=& H^2(K,\mathbf{G}_m) \stackrel{\text{def}}{\equiv} \text{Br}(K) \\ H^3(X,\mathbf{G}_m) &=& 0\\ H^3(X, j_\ast \mathbf{G}_m) &=& 0.\end{eqnarray}$$
I will compute the first of these and leave the rest as (great!) exercises for you. Let $$f : X \to \operatorname{Spec} \mathbf{F}_q$$ denote the structure map. Consider the Leray spectral sequence $$E_2^{i,j} := H^i(\operatorname{Spec} \mathbf{F}_q, R^jf_\ast \mathbf{G}_m) \implies E_\infty^{i+j} := H^{i+j}(X, \mathbf{G}_m).$$
The $E_\infty^2$ term (namely $H^2(X,\mathbf{G}_m)$) admits a filtration $$0 \subseteq F^0 \subseteq F^1 \subseteq E_\infty^2$$ with $$\begin{eqnarray} F^0 &\simeq& E_\infty^{2,0} \\ F^1/F^0 &\simeq& E_\infty^{1,1} \\ E_\infty^2/F^1 &\simeq& E_\infty^{0,2}.\end{eqnarray}$$ We now compute each of these. First, we deal with the $E_\infty^{2,0}$ term. From the 2nd page of the Leray spectral sequence, we see that $E_3^{2,0} \simeq E_2^{2,0}/\text{im}(E_2^{0,1} \to E_2^{2,0}).$ But $E_2^{2,0} = H^2(\operatorname{Spec} \mathbf{F}_q, \mathbf{G}_m)$ since $f_\ast \mathbf{G}_m = \mathbf{G}_m$ (cohomology and base change!). This can be identified with the Brauer group of $\mathbf{F}_q$ which is zero by Wedderburn's little theorem. Hence $E_3^{2,0} \implies E_\infty^{2,0}$ and hence $F^0 = 0$.
Now we show that $F^1 = 0$. Again, it is sufficient to show that $E_3^{1,1} = 0$. We see that $E_3^{1,1} \simeq \ker(E_2^{1,1} \to E_2^{3,0})$. However, the Galois cohomology of a finite field in odd degree is zero, so it is sufficient to show that $E_2^{1,1} = H^1(\operatorname{Spec} \mathbf{F}_q, R^1 f_\ast \mathbf{G}_m) = 0$. To this end, identify $R^1f_\ast \mathbf{G}_m$ with the Picard functor of $X/\mathbf{F}_q$, and consider the exact sequence $$0 \to \text{Pic}^0_{X/\mathbf{F}_q} \to \text{Pic}_{X/\mathbf{F}_q} \to \mathbf{Z} \to 0,$$ where the last map sends a line bundle to its degree. Taking cohomology, we have an exact sequence $$H^1(\operatorname{Spec} \mathbf{F}_q, \text{Pic}^0_{X/\mathbf{F}_q}) \to H^1(\operatorname{Spec} \mathbf{F}_q, \text{Pic}_{X/\mathbf{F}_q}) \to H^1(\operatorname{Spec} \mathbf{F}_q, \mathbf{Z}).$$ The left term is zero by Lang's theorem. The right term is zero because it can be identified with continuous homomorphisms from $\text{Gal}(\overline{\mathbf{F}}_q/\mathbf{F}_q)$ to $\mathbf{Z}$, of which there are none since the former is profinite while the latter discrete. This shows the middle term is zero, completing the proof that $E_\infty^{1,1} = 0$.
Finally, to show that $E_\infty^{2,0}$ is zero, observe that $R^2f_\ast \mathbf{G}_m$ is zero. The reason is because its stalk at the geometric point $\overline{x} : \operatorname{Spec} \overline{\mathbf{F}}_q \to \mathbf{F}_q$ is the Brauer group of $X \times \overline{\mathbf{F}}_q$. This is zero since it injects into the Brauer group of the function field which is zero by Tsen's theorem.