a) $(a+\frac{1}{a})^2=a^2+\frac{1}{a^2}+2 \implies (a+\frac{1}{a})^2-2=a^2+\frac{1}{a^2}$ Since $a+\frac{1}{a}$ is an integer, $(a+\frac{1}{a})^2$ is also an integer. Thus, $a^2+\frac{1}{a^2}$ is expressed as the difference of two integers, meaning that it must be an integer as well.

For (b) just use strong induction; base case can be $n = 2$ which was proved in (a), and then for the inductive step, we have $\left(a+\frac{1}{a}\right)\left(a^{n-1}+\frac{1}{a^{n-1}}\right) - \left(a^{n-2}+\frac{1}{a^{n-2}}\right) = a^n+\frac{1}{a^n}$, but the LHS is clearly an integer, and thus the RHS is as well.