Answer
Proofs below using a) contraposition and b) contradiction for the statement: if $n$ is an integer and $3n+2$ is even, then $n$ is even.
Work Step by Step
a) To prove the statement by contraposition, assume $n$ is an odd integer. We want to see whether $n^3+5$ is odd or even. If it is even, our theorem is proved. $n^3+5=n*n*n+5$. Because $n$ has no factors of 2, $n*n*n$ does not either and thus ends in the digit 1, 3, 5, 7, or 9. 5=4+1, and this 1 combines with 1, 3, 5, 7, or 9 to obtain 2, 4, 6, 8, or 0. Plus 4, these possible final digits are 6, 8, 0, 2, or 4, all of which mean the sum $n^3+5$ is even. Our theorem is proved.
b) Let $n\in \mathbb{Z}$ and $n^3+5$ be odd. Assume for contradiction that $n$ is odd. Then $n=n_e+1$, where $n_e$ is even. $n^3+5=(n_e+1)^3+5=n_e^3+3n_e^2+3n_e+1+5$, using the binomial expansion formula. The sum of the first three terms is even, because powers of an even number stay even; the sum of this even number and 6 is even. This contradicts our assumption that $n^3+5$ is odd, so our theorem is proved.
