Distinguished triangle and hypercohomology Let $X$ be a smooth geometrically integral variety over a number field $k$, we have an exact sequence of complexes of $\mathrm{Gal}(\bar{k}/k)$-modules
$$0 \rightarrow [\bar{k}^* \rightarrow 0] \rightarrow [\bar{k}(X)^* \rightarrow \mathrm{Div}(\bar{X})] \rightarrow [\bar{k}(X)^*/\bar{k}^* \rightarrow \mathrm{Div}(\bar{X})] \rightarrow 0.$$
As usual, the complexes in the middle (called 1-motives) are placed in degrees $-1$ and $0$. Taking hypercohomology groups, Lemma 2.1 of Local-global principles for 1-motives by Harari and Szamuely gives us the isomorphism $$\mathbf{H}^1(k,[\bar{k}(X)^* \rightarrow \mathrm{Div}(\bar{X})]) \cong \mathrm{Br}_1(X),$$
this is the algebraic Brauer group of $X$. I have two questions regarding its proof:
(1) It says that we have a distinguished triangle
$$\bar{k}(X)^* \rightarrow \mathrm{Div}(\bar{X}) \rightarrow [\bar{k}(X)^* \rightarrow \mathrm{Div}(\bar{X})] \rightarrow \bar{k}(X)^*[1],$$
but how do we check which exact triangle is this isomorphic to? This triangle itself is clearly not exact.
(2) The result seem to follow directly from the fact that the permutation module $\mathrm{Div}(\bar{X})$ has trivial $H^1$ (Galois cohomology), but I can't seem to see why.
 A: After referring to some sources, I shall try to provide an answer my own question.
With reference to the last paragraph of the answer by Dan Petersen to this question What is a triangle?, we apply the (hyper)cohomology functor $H^i(k,-)$ to the distinguished triangle given in question (1). This gives us the long exact sequence
$$H^1(k,\mathrm{Div}(\bar{X})) \rightarrow \mathbf{H}^1(k,[\bar{k}(X)^\times \rightarrow \mathrm{Div}(\bar{X})]) \rightarrow H^2(k,\bar{k}(X)^\times) \rightarrow H^2(k,\mathrm{Div}(\bar{X})).$$
Since $\mathrm{Div}(\bar{X})$ is a permutation module (indeed $\mathrm{Div}(\bar{X})$ is a free abelian group and it has a $\mathbb{Z}$-basis permuted by $\Gamma_k :=\mathrm{Gal}(\bar{k}/k)$), it is the sum of modules of the form $\mathbb{Z}[\Gamma_k/H]$ where $H$ is a subgroup of $\Gamma_k$. By Shapiro's lemma, for $i\geq 1$, we have
$$H^i(\Gamma_k,\mathbb{Z}[\Gamma_k/H]) \cong H^i(H,\mathbb{Z}) = 0.$$
Hence $H^1(k,\mathrm{Div}(\bar{X}))=0$ and so the middle map above is injective. The kernel of the last map is identified as $\mathrm{Br}_1(X)$ as seen in Diagram (4.17) on page 72 of Skorobogatov's book Torsors and rational points. Hence we obtain the desired isomorphism.
