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!
Here is a completely elementary example which shows that group cohomology is not empty verbiage, but can solve a problem ("parametrization of rational circle") whose statement has nothing to do with cohomology.
Suppose you somehow know that for a finite Galois extension $k\subset K$ with group $G$ the first cohomology group $H^1(G,K^*)$ is zero : this is the homological version of Hilbert's Theorem 90 ( you can look it up in Weibel's book on homological algebra, pages 175-176).
If moreover $G $ is cyclic with generator $s$, this implies that an element of $K$
has norm one if and only if it can be written $\frac{a}{s(a)}$ for some $a\in K$.
Consider now the quadratic extension $k=\mathbb Q \subset K=\mathbb Q(i)$ with generator $s$ of $Gal(\mathbb Q (i)/\mathbb Q)$ the complex conjugation.The statement above says that $x+iy\in \mathbb Q(i)$ satisfies $x^2+y^2=1$ iff
$x+iy=\frac{u+iv}{s(u+iv)}=\frac{u+iv}{u-iv}=\frac{u^2-v^2}{u^2+v^2}+i\frac{2uv}{u^2+v^2}$ for some $u+iv\in \mathbb Q (i)$ .
So we have obtained from group cohomology the well-known parametrization for the rational points of the unit circle $x^2+y^2=1$
$$x=\frac{u^2-v^2}{u^2+v^2}, \quad y=\frac{2uv}{u^2+v^2}$$.
Best Answer
To any ring $R$, we can associate the bounded derived category $D^b(R)$. The full subcategory $D^{perf}(R)$ is spanned by bounded complexes of projectives. If $R$ is self-injective, e.g. $R=k[G]$ the group-algebra of a finite group $G$ over a field $k$, then the Verdier quotient $D^b(R)/D^{perf}(R)$ is equivalent to the stable module category $\underline{mod}(R)$. If $R=k[G]$, and $M$ is an $R$-module, the Tate cohomology groups are
$$H^n(G,M)=\underline{mod}(R)(k,M[n]),\quad n\in\mathbb Z,$$ where $k$ is, as usual, the trivial $R$-module. The identification of the quotient with the stable module category is the non-trivial part of the story, and is due to Rickard ("Derived categories and stable equivalence").