A photomultiplier works by ejecting an electron every time a photon hits, and collecting it after it has been accelerated a great length, to knock out more electrons from more plates, and these are accelerated, and so on, until you get a macroscopic current you can measure. The noise comes from random electrons getting ejected from the plate anyway.
The answer is 1. Doing 2,3,5 increases the signal to noise, and doing 4 does nothing to signal to noise, but probably makes the multiplier stop working. The photomultiplier acts by ejecting electrons when photons hit, and decreasing the temperature will prevent electrons from getting ejected thermally randomly.
Higher voltage will lead to more thermal nucleation, since it will decrease the distance an electron has to go before it is knocked off the plate. Radioactive source is just stupid, you are introducing more noise. A lower "work function" (it's not a function of anything, it's the ionization energy) will mean the electrons have an easier time getting off the metal, so more random electrons ejected.
Option 4 is the only one that requires thinking. It will prevent electrons from reaching the collecting point, but if it doesn't affect the photo-plate it will just change the energy with which the ejected electrons strike, without changing the number of electrons of each type (photon emitted, or thermally emitted). This means it won't affect signal to noise. But it will probably increase signal to noise, because the photomultiplier is already tuned for good signal to noise, so reducing the voltage on the first collection point will perhaps reduce the multiplication to where you won't detect a signal.
Generally these questions are 40 years out of date, and the physics GRE is a joke. I had no idea what a "dynode" is, I guessed from context.