A charged black hole does produce an electric field. In fact, at great distances (much larger than the horizon), the field strength is $Q/(4\pi\epsilon_0 r^2)$, just like any other point charge. So measuring the charge is easy.
As for how the electric field gets out of the horizon, the best answer is that it doesn't: it was never in the horizon to begin with! A charged black hole formed out of charged matter. Before the black hole formed, the matter that would eventually form it had its own electric field lines. Even after the material collapses to form a black hole, the field lines are still there, a relic of the material that formed the black hole.
A long time ago, back when the American Journal of Physics had a question-and-answer section, someone posed the question of how the electric field gets out of a charged black hole. Matt McIrvin and I wrote an answer, which appeared in the journal. It pretty much says the same thing as the above, but a bit more formally and carefully.
Actually, I just noticed a mistake in what Matt and I wrote. We say that the Green function has support only on the past light cone. That's actually not true in curved spacetime: the Green function has support in the interior of the light cone as well. But fortunately that doesn't affect the main point, which is that there's no support outside the light cone.
Let's consider the simplest kind of black hole, which is one with no rotation or electric charge. The quick answer to the question is "yes, the mass is entirely within the Schwarzschild radius", but I would like to elaborate a little to explain what we are saying.
In the case of a more ordinary astronomical object, such as a planet or a star, one can have both the object itself and other things in orbit around it. The same is true for a black hole: there can be other things in orbit around it. So then the question arises, how in practice do we distinguish which mass belongs to the black hole, and which mass belongs to other things in orbit? This question is easy to answer for stable orbits, where the orbiting material is well outside the Schwarzschild radius. But when material is not in a stable orbit it may move towards the event horizon (at the Schwarzschild radius) and, in a time which will depend a lot on which observer is considered, it will pass the event horizon. After that the infalling mass soon reaches the singularity. We do not know the details of what transpires in the close vicinity of the central singularity, but the large scale picture is that the mass of the black hole has then grown a little, which we can tell by measuring the orbits of anything else still in orbit, or by gravitational lensing or other methods.
The horizon of a black hole is quite a subtle place because it is not just that the force of gravity gets to be very strong, it is more a case of the direction of time itself being "steered" in the inwards direction, so that once having passed the horizon, any object falling in cannot come out again just as surely as it cannot travel back to last week while going forward in time. This special feature applies to the horizon, and it means that right at the horizon there is a sort of cut-off: nothing can possibly hold an object fixed there. The object has to be either outside, or falling in and then it must keep on falling just as it cannot help moving forward in time.
Best Answer
I will approach this question theoretically, although I feel the intuition follows nicely. If we talk about Kerr black holes - rotating black holes described by their mass and angular momentum, with no additional parameters such as charge etc. - then you can show that the radius of the event horizon is given by
$\boxed{r=M + \sqrt{M^2-a^2}}$
where $a=\frac{J}{M}$.
(This value of $r$ is found by finding where the Kerr metric blows up; hence event horizon. In fact, finding where the metric blows up involves solving a quadratic equation, so we get two values of $r$ and in Kerr black holes we therefore have two event horizons; unlike in Schwarzschild black holes.)
Regarding your first point about maximum angular momentum, if we set $G=1$ and $c=1$, the maximum angular momentum you stated is given by $a=M$ and if we plug this into our equation for $r$ above we see that we have
$r=M$.
We know that the radius of the event horizon in a Schwarzschild black hole (no rotation) is $r=2M$. So therefore we can see that at maximum angular momentum, the radius of the event horizon is half of what it would be if the black hole weren't spinning.
To this end, we can also see that at zero angular momentum, $a=0$, we have
$r=2M$
which is what we want as at zero angular momentum we of course should have the Schwarzschild radius.
Using the boxed equation for $r$ at the top, it's easy to test out different values of $a$ to see what happens to the event horizon. For example, this equation alone is sufficient to show that for $a>M$ we don't have an event horizon, in which case we have what is a called "Fast Kerr" which is just a singularity with no event horizon.