No, the reasoning is a bit faulty - I think you've mixed up where the "not" goes. You can tell it's faulty because it's true that if mn is even then m is even or n is even.
You wanted "mn even => m even or n even" has contrapositive "m odd and n odd => mn odd".
If you're asked to prove that mn is even implies m even or n is even, I think you need a special case of Bézout's Theorem,which states that if hcf(m,n) = k then am+bn = k has solutions. This adapts by:
If m is not even, then hcf(2, m) = 1 and hence am + 2b = 1 has solutions, and hence amn + 2bn = n has solutions. But mn is assumed to be even, and hence amn is, so the LHS is even, and hence the RHS is even, so n is even. Hence at least one of m and n is even.
Your answer on slide 2 is fine.
Mx(t) is not negative because the integrand is positive (exponentials are positive).
Your Venn diagram is fine. P(C U D) = P(we get one of b,e,d,f); you can look at this as P(we don't get a or c).
The inverse-phi is exactly the inverse function of the phi function; that is, if phi(0) = 0.5, then phi^-1(0.5) = 0. You use the phi tables "in reverse" - that is, instead of looking at the row, and following along the columns to reach the right entry, you look for the entry in the table, then find which row/column it lies in.