Can you give a photon mass? I know this may sound weird. But could a photon be given mass? And if so what would the effects be?
 A: As far as I understand, if you gave a photon mass, the immediate consequence would be that photons wouldn't travel at speed $c$, and instead we would need to consider a new limiting speed for light, less than $c$. Even if your question seems a bit odd, it is pretty interesting, since the consideration of massive photons could give an explanation to dark energy.
A: If the photon mass arises from a Higgs mechanism then the photon mass can be switched on and off using an external magnetic field, as explained in this article. Above a critical magnetic field strength the Higgs vacuum expectation value will vanish and the photon becomes massless. 
A: Define mass. If you mean "rest mass", then the answer is no, the rest mass of a photon is zero, otherwise it couldn't propagate at light speed.
If you however mean "gravitational mass", then the answer is: It already has "mass" due to the infamous mass-energy equivalence $E=mc^2$ since a photon's energy equals $E=hf$ (where $h$ is Plack's constant and $f$ is the photon frequency). So a photon's gravitational mass is
$$m = \frac{hf}{c^2}.$$
But as the other answers already point out, the frequency and therefore "mass" depends on the reference frame! Also I am not aware of any experiments that had enough energy in a photon field to actually generate an observable gravitational influence...
A: If you give mass to a photon, one major consequence would be the loss of gauge invariance. 
The Lagrangian for electromagnetism,
$$\mathcal{L}\ =\ -\frac{1}{4}F_{\mu\nu}F^{\mu\nu}$$
is invariant under the gauge transformation
$$A_\mu\ \rightarrow\ \ A_\mu \ +\ \partial_\mu \Lambda $$
But, if you add a mass term, $m^2A_\mu A^\mu$ to the Lagrangian, it won't respect this symmetry anymore.
A: A "photon" propagating in an optical medium can be ascribed a nonzero rest mass. I put "photon" in quotes because light in a medium is not, strictly speaking, pure photons but a quantum superposition of excited electromagnetic field and matter quantum states.
In a medium with refractive index $n=1.5$, my calculation here reckons the rest mass of a quantum of this superposition to be:
$$m_0 = \frac{E}{c^2}\sqrt{1-\frac{1}{n^2}}$$
For $n=1.5$ (common glasses like window panes or N-BK7 - microscope slide glass) at $\lambda = 500\rm\,nm$, we get, from $E=h\,c/\lambda$, $m_0=3.3\times 10^{-36}{\rm kg}$ or about 3.6 millionths of an electron mass.
This take on your question is perhaps a little different from what your original question asks for, and is not to be confused with the assignment of a rest mass to the "pure photon" as discussed in the other answers. The assignment of nonzero rest mass to the photon replaces Maxwell's equations with the Proca equation, whose most "startling" characteristic is screening, i.e. photon fields would dwindle exponentially with distance from their sources and light as we know it could not propagate through the universe. 
