The written solutions follow below. I created a video to explain each step, but the recorded video answer appears to be in beta testing.
In problem one, we notice that the third premise happens to be the "if" of the "if-then" in the first premise. If J or R, then not C. Therefore, not C, therefore, if U then P. Easy as pie.
Problem One
1. (J v R) ⊃ ~C
2. C v ( U ⊃ P)
3 . J v R / U ⊃ P
4. ~C MP 1, 3
5. U⊃P DS 2, 4
This one is a little more sophisticated, but not difficult. Line 4 is the antecedent of the conditional in line 2. By Modus Ponens (MP), we learn that B. But if B, then not not B by the rule of double-negation. Therefore, not-not U, by Modus Tollens in line 3 -- and not-not U is also known as "U."
But then, if U, and if B (back on line 5), then we can start rolling up the first premise with modus ponens, one step at a time. We ultimately find our way to our conclusion -- not N.
Problem Two
1. U ⊃ [ U ⊃ ( B ⊃ ~ N)]
2. ( E v ~ J) ⊃ B
3. ~ U ⊃ ~ B
4. E v ~ J / ~ N
5. B MP 2, 4
6. ~~B DN 5
7. ~~U MT 3, 6
8. U DN 7
9. U ⊃ ( B ⊃ ~ N) MP 1, 8
10. B ⊃ ~ N MP 8,9
11. ~N MP 5, 10
