The front velocity, defined as the propagation speed of the point where the field first differs (by any arbitrarily small amount) from exactly zero, is always no greater than $c$. (In fact, the front velocity is always exactly equal to $c$, no greater or less.)
The problem here is that a Gaussian pulse extends infinitely in both directions, so it simply does not have a "front" to speak of. Of course the amplitude decays super-exponentially on both sides, but that doesn't matter. There is no causality problem with the pulse emerging arbitrarily early in time, because your input pulse made the field start changing long before that.
As a thought experiment, let's imagine we have a button we can push to start the input pulse going. If any trace of the output pulse comes out before a signal at $c$ has a chance to propagate from when and where the button was pushed, that's a causality violation. But that is impossible, for the following reason:
Since a theoretically perfect Gaussian pulse has no finite start time, but has a nonzero amplitude at arbitrarily early times, it's impossible to create such a perfect Gaussian pulse by pushing a button. Of course, you can get arbitrarily close to perfection, but there will always be some distortion that gets worse and worse as you try to make the pulse fire "faster", that is, try to make a shorter separation in time between the button push and the maximum of the pulse. The theorem that the medium has a causal response to the field guarantees that the response to this distortion will always interfere with the response to the perfect Gaussian to cancel out exactly at times earlier than a signal traveling at $c$.
Of course, in a real experiment, the "button push" (actually some electronic signal to the machine that creates the pulse) happens a relatively long time before any trace of the output pulse is detected.
Here's a great book chapter I just found about this: http://books.google.com/books?id=kE8OUCvt7ecC&pg=PA26 It has lots more math than my answer (though not too high a level) and might clear up a lot of things.
You also ask about single photons, but that opens up a huge can of worms I can't really get into (not least because I don't understand it well enough myself). Let me just say that there is always a minimum amount of noise in any mode / degree of freedom of the electromagnetic field, which is equivalent to half a photon. You could make a pulse of the right shape that's so weak there's only a single photon in it (I'm sure people do things like that all the time), but the problem is if you try to "announce" too early that you've detected that photon, you'll be wrong such a large fraction of the time that you can show statistically that no information is being transmitted. It's really difficult to do quantum-limited measurement in the first place (because you have to severely limit the back-reaction of the measurement apparatus on the field you're measuring), and if you try to do it too fast it becomes literally impossible.
Best Answer
In industry, cutting lasers are not selected based on their wavelength. They are selected based on their cost, which is closely related to energy efficiency. Carbon dioxide lasers, which are infrared lasers, are used because they have the best energy efficiency.
"Clear" glass will absorb some portion of the infrared spectrum, as will any other solid. So some infrared radiation, depending on the wavelength, will cut the glass better than visible radiation, which is not absorbed. The exact mechanism of cutting can vary.
It is also important to consider the geometry of the beam of radiation; owing to the diffraction limit, this is not independent of the wavelength.