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.
$\newcommand\F{\mathbf{F}}$
$\newcommand\Z{\mathbf{Z}}$
$\newcommand\SL{\mathrm{SL}}$
Because of the existence of the Weil paring,
elliptic curves with such a subgroup only exist when
$p \equiv 1 \mod \ N$.
Let $S_N$ denote the set of elliptic curves over $\F_p$ such
that $E[N]$ is defined over $\F_p$.
It will be slightly easier to assume that $N \ge 3$. In this case,
$Y(N)$ is a fine moduli space, and an $\F_p$-point on $Y(N)$
corresponds to
a pair $(E,\alpha:E[N] \simeq \Z/N\Z \times \Z/N \Z)$ defined over
$\F_p$. Given an elliptic curve $E \in S_N$,
how many points does it contribute to $Y(N)$? For a curve $E$
whose automorphism group is $\Z/2\Z$, We see that out of
the $|\SL_2(\Z/N\Z)|$ possible choices of $\alpha$
(technical remark, we have fixed a Weil pairing so that $Y(N)$ is
connected), $(E,\alpha) \simeq (E,\alpha')$ only if $\alpha' = \alpha$
or $\alpha' = [-1] \alpha$. Thus
$E$ contributes $|\SL_2(\Z/N\Z)|/2$ points to $Y(N)(\F_p)$.
In general, $E$ may have slightly more
automorphisms, and we deduce that (for $N \ge 3$):
$$|Y(N)(\F_p)| = |\SL_2(\Z/N\Z)| \sum_{E \in S_N}
\frac{1}{|\mathrm{Aut}(E)|}.$$
Note that the quantity on the right is very close to
$|\SL_2(\Z/N\Z)| \cdot |S_N|/2$, one only has to worry about the
elliptic curves with $j = 0$ or $j = 1728$, and this can be done by hand
if one wants to cross all the i's and dot all the t's.
Suppose that $X(N)$ has $c_N$ cusps and genus $g_N$ (there are some
explicit slightly unpleasant formulas for these numbers, which can
be found (for example) in Shimura's book. All the cusps
are defined over $\F_p$ (with $p \equiv 1 \mod N$) so by the
Riemann hypothesis for finite fields,
$$|Y(N)(\F_p) - (1+p) + c_N| = |X(N)(\F_p) - (1+p)| \le
2 g_N \sqrt{p}.$$
If $g_N = 0$ (which only happens if $N \le 5$), this leads to an
exact formula for $|S_N|$.
In general, at least for large $p$, we see that
$$|S_N| \sim \frac{2p}{|\SL_2(\Z/N \Z)|}.$$
To make this all completely explicit for $N = 3$ (for example),
one gets, presuming I have not made a horrible computational error which is quite possible:
$$S_3 = \begin{cases} (p+11)/12, & p \equiv 1 \mod \ 12, \\\
(p+5)/12, & p \equiv 7 \mod \ 12. \end{cases}$$
(note that $p \equiv 1 \mod 3$):
Of course, "exact formulas" will only exist for $N \le 5$.
Some related and slightly more difficult counting problems are also
nicely explained by Lenstra here (See 1.10):
https://openaccess.leidenuniv.nl/bitstream/1887/3826/1/346_086.pdf
Best Answer
3) Yes. They have the same characteristic polynomial of Frobenius acting on the Tate module, hence the same number of points.
4) Depends on how you want to reduce mod p. Certainly if they are isogenous with the same conductor, then they will have the same reduction type: split / non-split multiplicative reduction or additive reduction, corresponding to p-1, p+1, or p points. (Or 1 more, counting the singular point.)
However another reasonable way to count points over the reduction includes the number of components in the component group of the Neron model. This is not preserved by isogeny.