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?
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.
Best Answer
The remark added to the question shows that the kernel of $Ш(E/F) \to Ш(E/L)^G$ is finite where $G$ is the finite Galois group of $L/F$. $\DeclareMathOperator{\coker}{coker}$
Here is an argument why the cokernel is finite. It may be too complicated. Let $p$ be a prime and let us show that the cokernel on the $p$-primary part is finite (and that it is trivial for all $p$ that do not divide $\lvert G\rvert$). Let $\alpha\colon S(E/F) \to S(E/L)^G$ where $S$ stands for the $p$-primary Selmer group in $H^1\bigl(F, E[p^{\infty}]\bigr)$. Consider the exact sequence $$ 0\to \Bigl( E(L) \otimes \mathbb{Q}_p/\mathbb{Z}_p \Bigr)^G \to S(E/L)^G \to Ш(E/L)[p^{\infty}]^G \to H^1\bigl( G, E(L) \otimes \mathbb{Q}_p/\mathbb{Z}_p\bigr) $$ Note that the last term is isomorphic to $H^2\bigl(G,E(L)\otimes \mathbb{Z}_p\bigr)$, which is finite. Comparing that sequence with the sequence with $S(E/F)$ in the middle, shows that the cokernel on Ш lies in an exact sequence between $\coker(\alpha)$ and $H^2\bigl(G,E(L)\otimes \mathbb{Z}_p\bigr)$.
Let $S$ be the finite set of places containing all places at $\infty$, all places of bad reduction and all places above $p$. Let $G_S(L)$ be the Galois group of the maximal extension of $L$ unramified outside $S$. Next the exact sequence $$ 0\to S(E/L)^G \to H^1\bigl(G_S(L), E[p^{\infty}]\bigr)^G \to \Bigl(\bigoplus_{w \in S} H^1\bigl(L_w, E\bigr)[p^{\infty}]\Bigr)^{G} $$ Again we compare it with the similar sequence defining $S(E/F)$. This shows that there is a map $\beta\colon\coker(\alpha)\to H^2\bigl(G,E(L)[p^{\infty}]\bigr)$. The target of $\beta$ is finite; its kernel can be shown to be a subquotient of kernel of the restriction map on the right hand side: $$\bigoplus_{v \in S_F} H^1\bigl( G_w, E(L_w)\bigr)[p^{\infty}]$$ where $S_F$ is the set of places below $S$ and for each $v$ we choose one $w$ above $v$. As this is a finite sum of finite groups, this is also finite.
For the second question. Since $Ш(E/L)\subset H^1\bigl(L,E)$, your limit is natrually a subgroup of the limit of these $H^1$, i.e of $H^1(\bar{L},E)=0$.