You need to be a lot more careful when you use the phrase red-shift, due to how frequency is measured in general relativity. Roughly speaking, a photon is characterised by its wave vector $k$, which is a light-like four vector. The frequency measured by an observer is $g(\tau,k)$ where $g$ is the metric tensor, and $\tau$ is the unit vector tangent to the observer's world-line.
A little bit of basic Lorentzian geometry tells you that for any given photon $k$, instantaneously there can be observers seeing that photon with arbitrary high and arbitrarily low frequency.
So: start with your space-time. Fix a point on the event horizon. Fix a photon passing through that space-time event. For any frequency you want to see, you can choose a time-like vector at that space-time event that realises that frequency. Now, since the vector is time-like and the event horizon is null, the geodesic generated by that vector must start from outside the event horizon and crosses inside. Being a geodesic, it represents a free fall. So the conclusion is:
For any frequency you want to see, you can find a free falling observer starting outside of the black hole, such that it crosses the event horizon at the given space-time event and observes the frequency you want him to see.
So you ask, what is this whole business about gravitational red-shift of Schwarzschild black holes? I wrote a longer blog post on this topic some time ago and I won't be as detailed here. But the point is that on the Schwarzschild black holes (and in general, on any spherically symmetric solution of the Einstein's equations), one can break the freedom given by local Lorentz invariance by using the global geometry.
On Schwarzschild we have that the solution is stationary. Hence we can use the time-like Killing vector field* for the time-translation symmetry as a "global ruler" with respect to which to measure the frequency of photons. This is what it is meant by "gravitational redshift" in most textbooks on general relativity (see, e.g. Wald). Note that since we fixed a background ruler, the frequency that is being talked about is different from the frequency "as seen by an arbitrary infalling observer".
(There is another sense in which redshift is often talked about, which involves two infalling observers, one "departing first" with the second "to follow". In this case you again need the time-translation symmetry to make sense of the statement that the second observer "departed from the same spatial point as the first observer, but at a later time.)
It turns out, for general spherically symmetric solutions, there is this thing called a Kodama vector field, which happens to coincide with the Killing vector field on Schwarzschild. Outside of the event horizon, the Kodama vector field is time-like, and hence can be used as a substitute for the global ruler with respect to which to measure red-shift, when the space-time is assumed to be spherically symmetric, but not necessarily stationary. Again, this notion of redshift is observer independent. And it has played important roles (though sometimes manifesting in ways that are not immediately apparently related to red-shift, through choices of coordinates and what-not) in the study of dynamical, spherically symmetric gravitational collapse in the mathematical physics literature.
To summarise:
If you just compare the frequency of light measured (a) at its emission at the surface of the star in the rest frame associated to the collapse and (b) by an arbitrary free-falling observer, you can get basically any values you want. (Basically because the Doppler effect depends on the velocity of the observer, and you can change that to anything you like by choosing appropriate initial data for the free fall.)
One last comment about your last question:
You asked about what happens in the interior of the black hole. Again, any frequency can be realised by time-like observers locally. The question then boils down to whether you can construct such time-like observers to have come from free fall starting outside the black hole. By basic causality considerations, if you start with a time-like vector at a space-time event inside the black hole formed from gravitational collapse, going backwards along the time-like geodesic generated by the vector you will either hit the surface of your star, or exit the black hole. Though precisely how the two are divided depends on the precise nature of the gravitational collapse.
I should add that if you use the "global ruler" point of view, arguments have been put forth that analogous to how one expects red shift near the event horizon, one should also expect blue shifts near any Cauchy horizon that should exist. This has been demonstrated (mathematically) in the Reissner-Nordstrom (and similar) black holes. But as even the red-shift can sometimes run into problems (extreme charged black holes), one should not expect the statement about blue shifts near the Cauchy horizons to be true for all space-times.
Actually, it doesn't have the same mass, it has significantly less mass than its precursor star. Something like 90% of the star is blown off in the supernova event (Type II) that causes the black holes.
The Schwarzschild radius is the radius at which, if an object's mass where compressed to a sphere of that size, the escape velocity at the surface would be the speed of light $c$; this is given by
$$
r_s=\frac{2Gm}{c^2}
$$
For a 3-solar mass black hole, this amounts to about 10 km. If we measure the gravitational acceleration from this point,
$$
g_{BH}=\frac{Gm_{BH}}{r_s^2}\simeq10^{13}\,{\rm m/s^2}
$$
and compare this to the acceleration due to the precursor 20 solar mass star with radius of $r_\star=5R_\odot\simeq7\times10^8$ m, we have
$$
g_{M_\star}=\frac{Gm_\star}{r_\star^2}\simeq10^3\,{\rm m/s^2}
$$
Note that this is the acceleration due to gravity at the surface of the object, and not at some distance away. If we measure the gravitational acceleration of the smaller black hole at the distance of the original star's radius, you'll find it is a lot smaller (by a factor of about 7).
Best Answer
During a supernova, a star blasts away its outer layers; this actually reduces the mass of the star significantly.
Any star or planet has an escape velocity - the slowest an object must be traveling for it to escape the gravitational field of the star/planet. For Earth, this is 11.2 km/s. (Note that this value doesn't account for any atmospheric effects.) For a black hole, however, the escape velocity at the event horizon (the "edge," in some sense) of the black hole is the speed of light, $c = 300,000 \text{ km/s}$. For anything within the event horizon to escape the pull of a black hole, it must exceed the speed of light, a physical impossibility. There's a certain mass-dependent radius - the Schwarzschild radius - to which an object must shrink in order to become a black hole.
Newton's Law of Universal Gravitation, which you stated, doesn't apply in its standard form to light. Rather, you need to use Einstein's general relativity, which considers gravitational forces in a much different light than Newton did. However, Newton's Law of Gravitation can intuitively apply here: when a star collapses, $m$ does decreases. However, $r$ becomes much smaller, so the net effect of these changes is the creating of a stronger gravitational force on the surface of the remaining object.