How can a black hole evaporate instantly if speed of gravity is c?

[PeterDonis]

While the metric indicates the stretching ratio goes to infinity at the horizon, the actual distance integrated over the metric remains finite. Thus a waveform leaving the horizon at velocity c should reach an observer in finite time.

For a static black hole that doesn't evaporate, a light signal emitted from anywhere above the horizon (not *at* it) will reach an observer anywhere outside the horizon in some finite time. However, the finite time can be arbitrarily large; it can be made as large as you like by putting the emitter of the waveform closer and closer to the horizon. This is not a matter of the distance from the emitter to the observer; it's a matter of the light cones being tilted inward, so that the light takes longer to cover the distance than it would if the light cones were not tilted. One could say that the light was "moving slower than c", but I don't like that terminology because it invites a lot of confusion about whether GR is consistent with SR; in fact, the rule that "light always moves at c" simply can't be applied as-is in a curved spacetime, because there is no way to uniquely define "relative speed" for spatially separated objects. The rule in GR becomes "light always moves along the light cones", which has a unique, unambiguous definition.

For an evaporating hole, the location of the horizon changes, and the spacetime as a whole is not static, so things get more complicated; but the upshot, as I said in previous posts, is that you can no longer make the finite time arbitrarily large.

2) Am I to take it that the model of evaporation is an instantaneous, uniform, symmetric burst of radiation traveling at a local speed of c?

No. Black hole evaporation is a continuous process. The hole starts with some mass M, and very slowly radiates it away, continuously, until it's all gone.