First of all, A B and C are all isothermal ideal gas scenarios. For such situations delta U is always 0.

A) you're right, P is 0 so w is 0. So calculate q.

B) This is an irreversible process because the gas is expanding against a fixed constant pressure. I

w = -[integral(P dV)]

Now, since P is a contant external pressure, the integration is simple.

w = -P(delta V)

The simple trick here is to substitute for the final Pressure from the ideal gas equation.

C) This is a reversible process. It is the same form as above, but with a difference:

w = -[integral(P dV)] = -[integral(nRT/V dV)]

= -(nRT)ln(V2/V1)

notice that here, P is not constant. It is substituted for and integrated because the external pressure is slowly changing as a gas slowly expands in a reversible process.