um, yeah, well, i think you're on the right track.

first, looking at the choices, you know that you can eliminate A and D because I and II are opposite things, so both can't happen at the same time. In other words, the work is done by the surroundings ON the system, so both the work done by surroundings and the work done on the system should have the same sign. that's just how you can reason through eliminating some answers, but...

to the real thermodynamics! ok, so I don't know about you, but I find that the best way to think of work done and work lost is to think of the system as exerting a force upwards on the piston, while the surroundings exert a downward force on the system. now, if the gas is then compressed, that means that the work done by the surroundings is positive, because the piston went down (which is displacement in the **same** direction as the force exerted by the surroundings, so you get positive work). another way of saying this is that the work done ON the system is positive. at the same time, the work done BY (not ON) the system should be negative because the system exerts an upward force on the piston but it is still moving downwards (so displacement in the opposite direction of the force), so you get negative work.

so, we now know that II is correct, while I is incorrect.

What about 3?

ok then, if you have deltaU=q-w, well, remember that w is the work done by the system. so w is negative, and in an adiabatic process q=0. so deltaU will equal a positive value (negative of the negative work). Since deltaU=int(CvdT), if deltaU is positive and Cv is a constant, T must increase to give a positive deltaU.

sooo, temperature increases.

hope that helps.