Reflecting sunlight --- does this do any Work? Photons have no mass, and yet via radiation pressure they do exert a force on whatever they reflect off of. Therefore the object must also exert an equal and opposite force on the photon.
So if Work is force times distance, this must do some kind of work. But I don't know how to calculate it because I don't know what distance to use. I think the work might be very high because going at the speed of light, then suddenly reversing direction, would be a delta v of 2c requiring extreme acceleration and thus extreme force (actually, extreme F/m, but photon mass is zero...).
(I have to throw this in): Also, why wouldn't this be a perpetual motion machine?
 A: A stream of photons which is reflected by a perfect mirror doesn't, in spite of the radiation pressure, do any work on the mirror unless the mirror moves (possibly due to the radiation pressure). There is no transfer of energy from the photons to the mirror. This is similar to the pressure of gas molecules on the wall of a container, which is not moving. If the mirror moves with velocity v in the direction of the photon flux, the reflected photons experience a Doppler frequency reduction and thus a loss of energy which is imparted to the mirror.
A: Don't try and calculate it using forces, do it using energy and momentum instead. First conserve momentum:a number of photons strike a mirror on the earth and recoil. Technically they change their frequency by a tiny amount, but as far as conservation of momentum goes that is negligible. The mirror (plus the planet it is attached to) acquire an equal change in momentum in the opposite direction. From this you can calculate the change in velocity of the earth. 
Then using that velocity you can calculate the change in KE of the earth (in the frame or was initially at rest), which by conservation of energy is going to give you the loss in energy of the photon. And obviously you can figure out the change in frequency from that. 
It will be tiny. Really tiny. Which is why it is general ignored, because it almost never matters at all. But strictly speaking you are right that ignoring it means you are violating conservation of energy. Is just that doing it 'properly' makes no difference at all in most cases but is more work, do why not simplify the calculations and get the same result? 
A: Think about it like this: 
A photon travels towards the mirror (or solar sail!) with three-momentum $p$. Assuming perfect reflection, it will bounce off the mirror and go back the way it came, now with momentum $-p$. Thus, it will have transferred a momentum of $p - (-p) = 2p$ to the mirror.
