This question has two parts, the first part will be to obtain the desired exact sequence while the second will be to study it in the corresponding derived category and try to obtain it from there.

Let $X$ be a smooth geometrically integral variety over a number field $k$ with canonical morphism $\pi:X \rightarrow \mathrm{Spec}\,k$, I want to obtain the following exact sequence of $\mathrm{Gal}(\bar{k}/k)$-modules (all tensor products are over $\mathbb{Z}$):

$$\mu_\infty \rightarrow \bar{k}[X]^* \rightarrow \bar{k}[X]^* \otimes \mathbb{Q} \rightarrow H^1(\bar{X},\mu_\infty) \rightarrow \mathrm{Pic}(\bar{X}) \rightarrow \mathrm{Pic}(\bar{X})_{free} \rightarrow 0,$$

where $\bar{X} := X \times _k \bar{k}$, $\mu_\infty = \mathrm{colim}_n\mu_n$, and $\mathrm{Pic}(\bar{X})_{free}$ denotes the maximal free quotient of $\mathrm{Pic}(\bar{X})$.

A very natural approach will be to apply the etale cohomology functor $H^i(\bar{X},-)$ to the exact sequence

$$1 \rightarrow \mu_\infty \rightarrow \mathbb{G}_m \rightarrow \mathbb{G}_m \otimes \mathbb{Q} \rightarrow 1.$$

To see why this sequence is exact refer to the answer of my post A Kummer exact sequence involving $\mu_\infty$.

So everything is fine except for the surjective map $\mathrm{Pic}(\bar{X}) \rightarrow \mathrm{Pic}(\bar{X})_{free}$, which can be rewritten as

$$\mathrm{Pic}(\bar{X}) \rightarrow \mathrm{Ker}(H^1(\bar{X},\mathbb{G}_m) \otimes \mathbb{Q} \rightarrow H^2(\bar{X},\mu_\infty)).$$

**Question 1.** How do I show that $\mathrm{Pic}(\bar{X})_{free}$ is precisely the kernel written above?

One thought I have is that perhaps we can try to show that $H^2(\bar{X},\mu_\infty) = 0$, and then we would need to prove that $\mathrm{Pic}(\bar{X}) \otimes \mathbb{Q} \cong \mathrm{Pic}(\bar{X})_{free}$, which would not be surprising since we are sort of 'killing off' the torsion parts of the Picard group.

Now on to the next part of the question, we have that the inclusion $\mu_\infty \rightarrow \mathbb{G}_m$ induces the map

$$\varphi: \tau_{\leq 1}R\pi_*\mu_\infty \rightarrow \tau_{\leq 1}R\pi_*\mathbb{G}_m$$

in the category of bounded complexes of sheaves of discrete Galois modules. Let $D := \mathrm{Cone}(\varphi)$, thus we have a distinguished triangle

$$\tau_{\leq 1}R\pi_*\mu_\infty \rightarrow \tau_{\leq 1}R\pi_*\mathbb{G}_m \rightarrow D \rightarrow (\tau_{\leq 1}R\pi_*\mu_\infty)[1].$$

This would give rise to the exact sequence

$$0 \rightarrow \tau_{\leq 1}R\pi_*\mathbb{G}_m \rightarrow D \rightarrow (\tau_{\leq 1}R\pi_*\mu_\infty)[1] \rightarrow 0.$$

Let $h^i(A)$ denote the $i$-th cohomology group of the complex $A$, then we have a long exact sequence of cohomology groups

$$h^{-1}(D) \rightarrow h^{-1}((\tau_{\leq 1}R\pi_*\mu_\infty)[1]) \rightarrow h^0(\tau_{\leq 1}R\pi_*\mathbb{G}_m) \rightarrow h^0(D) \rightarrow h^0((\tau_{\leq 1}R\pi_*\mu_\infty)[1])$$ $$\rightarrow h^1(\tau_{\leq 1}R\pi_*\mathbb{G}_m) \rightarrow h^1(D) \rightarrow h^1((\tau_{\leq 1}R\pi_*\mu_\infty)[1]).$$

Here we only consider $h^i(D)$ for $i = -1,0,1$ because by definition, the cohomology is zero outside these degrees. It is well-known that since $X$ is smooth over $k$, $\tau_{\leq 1}R\pi_*\mathbb{G}_m$ is quasi-isomorphic to the complex $[\bar{k}(X)^* \rightarrow \mathrm{Div}(\bar{X})]$ in degrees 0 and 1. Thus one easily computes that $h^0(\tau_{\leq 1}R\pi_*\mathbb{G}_m) \cong \bar{k}[X]^*$ and $h^1(\tau_{\leq 1}R\pi_*\mathbb{G}_m) \cong \mathrm{Pic}(\bar{X})$. Also, we have

$$h^i((\tau_{\leq 1}R\pi_*\mu_\infty)[1]) = h^{i+1}(\tau_{\leq 1}R\pi_*\mu_\infty)$$

and so the last term of the above long exact sequence is 0.

**Question 2.** We want to show that this long exact sequence is precisely the one mentioned in the first part of the question. All we need to do is find an injective resolution for $\mu_\infty$, this will enable us to get an explicit presentation of $R\pi_*\mu_\infty$, but this is where I have no idea how to proceed.