I'm not sure what's going on with this question, but let me drop a few lines to summarize what the official position should be.
First off, I can't think about any situation these groups may occur beyond that of a Hochschild-Serre spectral sequence, so I hope group cohomology here is meant to be continuous group cohomology, and $\mathbf{Z}_{\ell}$-cohomology must then mean continuous étale cohomology.
The given answer (the accepted one) is incorrect.
In the functorial short exact sequences
$$0\to {\lim}^1 H^{i-1}(X_{\rm ét},\mathbf{Z}/(\ell^n))\to H^i\to \lim H^i(X_{\rm ét},\mathbf{Z}/(\ell^n))\to 0$$
where $H^i$ is either $H^i(X_{\rm proét},\underline{\mathbf{Z}}_{\ell})$ or Jannsen's continuous étale cohomology $H^i_{\rm cont}(X,\{\mathbf{Z}/(\ell^n)\})$ (since they agree for any $X$) the ${\lim}^1$ term vanishes as soon as $H^{i-1}(X_{\rm ét},\mathbf{Z}/(\ell^n))$ is finite, by Mittag-Leffler.
In particular, if $X$ is defined over a separably closed field and proper (as in the OP's assumptions), $H^i$ agrees with usual $\ell$-adic cohomology.
Will's example is the typical way to show that geometric $\ell$-adic cohomology is usually not a discrete Galois representation, so any argument trying to infer $H^i(\text{Gal}(k^{\rm sep}/k), H^j)$ is torsion for $i>0$ and $j\ge 0$ from discreteness of $H^j$ is wrong.
Also, it is wrong to say that the same argument as for geometric étale cohomology of abelian sheaves, showing the Galois action is discrete, works for proétale cohomology. One can define the action abstractly and the concrete way as in the accepted answer, but these two actions don't generally agree anymore. It boils down to the fact that evaluation at usual geometric points does not give a conservative family of fiber functors on the proétale topos.
Both questions asked by the OP have negative answer, regardless of what $H^i$ is meant to be, among the possibilities discussed here.
Maybe the OP was meaning to ask something different?
Consider $E:y^2=x^3-3$ over $\mathbb Q$. Then $E(\mathbb Q)$ is trivial, hence so is $E(\mathbb Q)\otimes\mathbb Q^\times$. On the other hand, over $L=\mathbb Q(\sqrt{-3})$ we have a point $P=(0,\sqrt{-3})\in E(L)$ which is mapped to its opposite by the complex conjugation. The same is true of $\sqrt{-3}\in L$, therefore $P\otimes\sqrt{-3}\in(E(L)\otimes L^{\times})^{\text{Gal}(L/K)}$, and we can check this element is nontrivial.
An analogous example for torsion-free parts is given by the point $P=(0,\sqrt{-2})\in E(L)$ for $E:y^2=x^3+x-2$ and $L=\mathbb Q(\sqrt{-2})$. $E(\mathbb Q)\cong\mathbb Z/2$ and $P$ is a point of infinite order, so $P\otimes\sqrt{-2}$ will be nontrivial in $E(L)_{tf}\otimes L^\times$.
I don't know any result for $H^1$ but I doubt there is anything as simple as Hilbert's 90. As a Galois representation, $E(L)$ can have quite nontrivial structure, even after tensoring with $\mathbb Q$, but I don't have an explicit counterexample at hand. (edit: actually see Chris Wuthrich's comment below though)
Best Answer
I think that all Galois cohomology groups $\mathrm{H}^i(\mathrm{Gal}_K,M)$ vanish for $i>0$. As you observed, it suffices to prove that $\mathrm{H}^i(\mathrm{Gal}_K/U,M^U)$ vanishes for all open normal subgroups $U$ of $\mathrm{Gal}_K$. Since $\mathrm{Gal}_K/U$ is a finite group, the Galois cohomology group $\mathrm{H}^i(\mathrm{Gal}_K/U,M^U)$ is torsion (via restriction/corestriction with the trivial subgroup, see Serre: Corps locaux, p. 138). In fact, all of its elements have order a divisor of $n=|\mathrm{Gal}_K/U|$. It suffices then to show that the multiplication-by-$n$ map $[n]$ on $M^U$ is bijective.
The map $[n]\colon M^U\rightarrow M^U$ is injective since $M$ is torsion free. In order to show that it is surjective choose $x\in M^U$ and consider the short exact sequence defining $M$ $$ 0\rightarrow A(\bar K)_{\mathrm{tor}}\rightarrow A(\bar K)\overset{\pi}{\rightarrow} M\rightarrow 0, $$ where $A(\bar K)_{\mathrm{tor}}$ is the torsion subgroup of $A(\bar K)$. Since $x\in M$ and $A(\bar K)$ is a divisible group, there is an element $y\in A(\bar K)$ such that $\pi([n]y)=x$, where $[n]$ also denotes the multiplication-by-$n$ morphism on the abelian group $A(\bar K)$. Let us prove that $\pi(y)\in M^U$. Choose $g\in U$. One has $$ \pi([n]gy)=\pi(g[n]y)=g\pi([n]y)=gx=x=\pi([n]y) $$ since $x\in M^U$. It follows that $[n](gy-y)\in\ker(\pi)$. Hence $[n](gy-y)$ is a torsion element of $A(\bar K)$. Then $gy-y$ is in $A(\bar K)_{\mathrm tor}$ as well. This means that $$ g\pi(y)-\pi(y)=\pi(gy-y)=0, $$ i.e., $g\pi(y)=\pi(y)$. Hence $\pi(y)\in M^U$. Since $[n]\pi(y)=\pi([n]y)=x$, the morphism $[n]\colon M^U\rightarrow M^U$ is surjective indeed.