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!
If all Tate modules (i.e., for all $\ell$) are isomorphic then they differ by
the twist by a locally free rank $1$ module over the endomorphism ring of one of
them. This is true for all abelian varieties but for elliptic curves we only
have two kinds of possibilities for the endomorphism ring; either $\mathbb Z$ or
an order in an imaginary quadratic field. In the first case there is only one
rank $1$ module so the curves are isomorphic. In the case of an order we get
that the numbe of twists is a class number.
Addendum: Concretely, we have that $\mathrm{Hom}(E_1,E_2)$ is a rank $1$ projective module over $\mathrm{End}(E_1,E_1)$ (under the assumption that the Tate modules are isomorphic) and then $E_2$ is isomorphic to $\mathrm{Hom}(E_1,E_2)\bigotimes_{\mathrm{End}(E_1)}E_1$ (the tensor product is defined by presenting $\mathrm{Hom}(E_1,E_2)$ as the kernel of an idempotent $n\times n$-matrix with entries in $\mathrm{End}(E_1)$ and $E_2$ is the kernel of the same matrix acting on $E_1^n$. Hence, given $E_1$ $E_2$ is determined by $\mathrm{Hom}(E_1,E_2)$ and every rank $1$ projective module appears in this way.
Addendum 1: Note that I was talking here about the $\mathbb Z_\ell$ (and not $\mathbb Q_\ell$ Tate modules. You can divide up the classification of elliptic curves in two stages: First you see if the $V_\ell$ are isomorphic (and there it is enough to look at a single $\ell$). If they are, then the curves are isogenous. Then the second step is to look within an isogeny class and try to classify those curves.
The way I am talking about here goes directly to looking at the $T_\ell$ for all $\ell$. If they are non-isomorphic (for even a single $\ell$ then the curves are not isomorphic and if they are isomorphic for all $\ell$ they still may or may not be isomorphic, the difference between them is given by a rank $1$ locally free module over the endomorphism ring. In any case they are certainly isogenous. These can be seen a priori as if all $T_\ell$ are isomorphic so are all the $V_\ell$ but also a posteriori essentially because a rank $1$ locally free module becomes free of rank $1$ when tensored with $\mathbb Q$.
Of course the a posteriori argument is in some sense cheating because the way you show that the curves differ by a twist by a rank $1$ locally free module is to use the precise form of the Tate conjecture:
$$
\mathrm{Hom}(E_1,E_2)\bigotimes \mathbb Z_\ell = \mathrm{Hom}_{\mathcal G}(T_\ell(E_1),T_\ell(E_2))
$$
which for a single $\ell$ gives the isogeny.
Note also that the situation is similar (not by chance) to the case of CM-curves. If we look at CM-elliptic curves with a fixed endomorphism ring, then algebraically they can not be put into bijection with the elements of the class group of the endomorphism ring (though they can analytically), you have to fix one elliptic curve to get a bijection.
Best Answer
Allow me to say something which is not so much an answer to this question as to a (very natural) question that I sense is coming in the future.
There are two possible pitfalls in the definition of "has complex multiplication" for abelian varieties over an arbitrary ground field $F$.
1) Let's first talk only about complex abelian varieties. There is a serious discrepancy between the terminology "has complex multiplication" as used in the classical literature (up to and including, say, some of Shimura's papers in the 1960s) and the way it is almost invariably used today.
Classically, an abelian variety over the complex numbers was said to "have complex multiplications" whenever its endomorphism ring was strictly larger than $\mathbb{Z}$. This is never the generic case, but starting in dimension $2$ it allows abelian surfaces with multiplication by an order in a real quadratic field (parameterized by Hilbert modular surfaces), by an order in an indefinite rational quaternion algebra (parameterized by Shimura curves), and so forth.
Nowadays one means something much more restrictive by "complex multiplication": one wants the endomorphism ring to be an order in a $\mathbb{Q}$-algebra whose dimension is large compared to the dimension, say $g$, of $A$. In the case of an abelian variety without nontrivial abelian subvarieties ("simple"), CM means precisely that the endomorphism ring is an order in a number field of degree $2g$: it then follows from this (not so obviously) that the number field is totally complex and has an index $2$ totally real subfield. For nonsimple abelian varieties, the generally agreed upon definition is that $\operatorname{End}(A)$ should contain an order in a commutative semisimple $\mathbb{Q}$-algebra of dimension $2g$. Such things do not vary in moduli, even in higher dimensions.
2) If you are interested in abelian varieties over an arbitrary ground field $F$ (let me concentrate on the case of characteristic 0), then you need to distinguish between the ring of endomorphisms which are rationally defined over $F$ and the ring of endomorphisms which are defined over an algebraic closure of $F$ (hence, it turns out, over any algebraically closed field containing $F$, i.e., one cannot acquire endomorphisms by passing from $\overline{\mathbb{Q}}$ to $\mathbb{C}$). If one does not specify, "having CM" means having CM over the algebraic closure.
As in the responses above, it is very often the case that you have to extend the ground field slightly in order to get all the endomorphisms rational over the ground field. E.g. if $E_{/\mathbb{C}}$ is any elliptic curve, it can be minimally defined over $\mathbb{Q}(j(E))$. If $E$ has complex multiplication (over $\mathbb{C}$), then $j(E)$ is a real [not necessarily totally real] algebraic number, from which it follows that $\operatorname{End}_{\mathbb{Q}(j(E))}(E) = \mathbb{Z}$.
In particular, the $\ell$-adic Galois representation of a "CM abelian variety" [using standard conventions as in 1) and 2) above] need not have abelian image, quite. It has "almost abelian image", meaning that once you make a finite extension of the ground field to make all the endomorphisms rational, then the image will be abelian. (Both M. Emerton's and my arguments in the one-dimensional case extend easily to this case, although I think his is more insightful.)
What I have said carries over to fields of positive characteristic, but the general picture of endomorphism algebras looks different, especially over finite fields: every abelian variety over a finite field has complex multiplication in the above sense.