For part (a), simply note that $\sqrt{n^2 + a} \in \mathbb{Q}$ iff $n^2 + a$ is a square, and notice that $a = x^2 - n^2 = (x-n)(x+n)$ clearly has finitely many solutions in $(x, n).$ As for part (b), notice that for sufficiently large $n$, $|n\sqrt{n^2 + a} - n^2 + \frac{a}{2}|$ approaches zero. Therefore, it suffices to show that for $n$ sufficiently large, there exists a constant $c$ for which $|n^2 + \frac{a}{2} - bm| \ge c$ whenever it's nonzero. However, as $n^2 + \frac{a}{2}$ always has fractional part $\frac12$ as $a$ is odd, and $bm$ is always at least $\frac{1}{2q}$ from any number with fractional part $\frac12$ (where $q$ denotes the odd denominator of $b$ in simplest form), we are done by simply taking $c = \frac{1}{2q}$. Now pick a suitable $C$ for the problem at hand by looking at cases where $n$ is small. $\square$