IMO, the scenario is closer to your (a). I'll sketch an explanation of the duality between $H^1(E,\mathbf{Z}_l)$ and the dual to the Tate module. We have $H^1(E,\mathbf{Z}_l)=\text{Hom}(\pi_1(E),\mathbf{Z}_l)$,
where that $\pi_1$ means etale fundamental group with base point the origin $O$ of $E$. Thus the isomorphism we really want is between $\pi_1(E)\otimes\mathbf{Z}_l$ and $T_\ell(E)$.
What is $\pi_1(E)$? In the topology world, we'd consider the universal cover $f\colon E'\rightarrow E$ and take $\pi_1(E)$ to be its group of deck transformations. Then $\pi_1(E)$ has an obvious action on $f^{-1}(O)$. If $E$ is the complex manifold $\mathbf{C}/L$ for a lattice $L$, this is just the natural isomorphism $\pi_1(E)\cong L$.
But in the algebraic geometry world, there is no universal cover in the category of varieties, so the notion of universal cover is replaced with the projective system $E_i\to E$ of etale covers of $E$. Then $\pi_1(E)$ is the projective limit of the automorphism groups of $E_i$ over $E$.
One nice thing about $E$ being an elliptic curve is that any etale cover $E'\rightarrow E$ must also be an elliptic curve (once you choose an origin on it, anyway); if $E\rightarrow E'$ is the dual map then the composition $E\rightarrow E'\rightarrow E$ is multiplication by an integer. So it's sufficient to only consider those covers of $E$ which are just multiplication by an integer. Since it's $\pi_1(E)\otimes\mathbf{Z}_l$ we're interested in, it's enough to consider the isogenies of $E$ given by multiplication by $l^n$.
What are the deck transformations of the maps $l^n\colon E\rightarrow E$? Up to an automorphism of $E$, they're simply translations by $l^n$-division points. And now we see the relationship to the Tate module: A compatible system of deck transformations of these covers is the exact same thing as a compatible system of $l^n$-division points. Thus we get the desired isomorphism. Naturally, it's Galois compatible!
In the end, we see that torsion points were tucked away in the construction of the etale cohomology groups, so it wasn't exactly a coincidence. Hope this helps.
Re the edit: I believe your best bet is to work locally. First of all, you didn't mention which Galois representation you wanted exactly; let's say you want the representation on $H^i$ of your variety for a given $i$. Let's assume this space has dimension $d$.
Step 1. For each prime $p$ at which your variety $V$ has good reduction, you can compute the local zeta function of $V/\mathbf{F}_p$ by counting points on $V(\mathbf{F}_{p^n})$ for $n\geq 0$. In this way you can compute the action of the $p^n$th power Frobenius on $H^i(V\otimes\overline{\mathbf{F}}_p,\mathbf{F}_5)$ for various primes $p$.
Step 2. Do this enough so that you can gather up information on the statistics of how often the Frobenius at $p$ lands in each conjugacy class in the group $\text{GL}_d(\mathbf{F}_5)$. In this way you could guess the conjugacy class of the image of Galois inside $\text{GL}_d(\mathbf{F}_5)$.
Step 3. Now your job is to find a table of number fields $F$ whose splitting field has Galois group equal to the group you found in the previous step. I found a table here: http://hobbes.la.asu.edu/NFDB/. You already know which primes ramify in $K$ -- these are at worst the primes of bad reduction of $V$ together with 5 -- and you can distinguish your $F$ from the other number fields by the splitting behavior your found in Step 1. Then $K$ is the splitting field of $F$.
A caveat: Step 1 may well take you a very long time, because unless your variety has some special structure or symmetry to it, counting points on $V$ is Hard.
Another caveat: Step 3 might be impossible if $d$ is large. If $d$ is 2 then perhaps you're ok, because there might be a degree 8 number field $F$ whose splitting field has Galois group $\text{GL}_2(\mathbf{F}_5)$. If $d$ is large you might be out of luck here.
You are free not to accept this answer because of the above caveats but I really do think you've asked a hell of a tough question here!
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?
Best Answer
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)