A zinc-copper battery is constructed as follows:

Perhaps I can get you started but that is it.
Since the concns are not standard, first calculate the Eo value for each half cell but BOTH as reductions.
For Zn^+ + 2e = Zn we have
E = Eo-(0.0592/n)*log[(Zn)/(Zn^+2)] =
-.763-(0.0592/2)*log[1/0.1)] = -0.793v. (check my work on this). This will be used as an oxidation reaction; therefore, reverse the equation and change the sign.
Eox = +0.793 v.

For the Cu, as a reduction.
E = Eo-(0.0592/n)*log[(Cu)/(Cu^+2)] =
+0.337 - (0.0592/2)*log(1/2.50) = 0.348v. Check this arithmetic, too.
Ered = 0.348 v.
Ecell = Eox + Ered = 0.793+0.348 = 1.14 v. Check this arithmetic (and chemistry).
The above is part a in DETAIL.
Post your work on the others if you get stuck and someone will help. Some hints:
amperes x seconds = coulombs.
96,485 coulombs will deposit one equivalent of a metal (1 equivalent = 1/2 mole for both Cu and Zn). For (b), you should start by calculating the amount of Zn that goes into solution and how much Cu^+2 comes out of solution, use those new concns to recalculate the half cell voltages just as I did in part (a), then calculate the new cell potential. Part (c) will follow from part (b). Follow the hint for part (d).