Okay, so this is what I did. How does it look? I found the final molarity, pH, and OH- (by finding H+ and re-arranging Kw).
0.03 mol/L * 0.01 L = 0.0003 mol
7.5 x 10-3 mol/L * 0.04 L = 0.0003 mol
+ = 0.0006 mol
0.0006mol / 0.05 L = 0.012 mol/L
pOH = -log(0.012mol/L) = 1.92
pH = 14.00 - 1.92 = 12.08
[H+] = 10-12.08 = 8.32 * 10-13
[OH-] = 1.00 * 10-14 / 8.32 * 10-13 = 1.2 * 10-2