a) The solution of part a) is almost obvious and its statement holds for any $b \in \mathbb{Q}$ and $a \in \mathbb{N}$. Indeed, if $n\sqrt{n^2+a} = bm$, then, since $b \in \mathbb{Q}$, the number $\sqrt{n^2+a}$ must be rational and therefore integer. Hence $n^2+a = k^2$ for some positive integer $k$. But the differences between consecutive perfect squares infinitely increase, there is only finite number of pairs $(n, k)$ of positive integers $n$ and $k$ such that for the given number $a$ holds $n^2 - k^2 = a$.

b) Let $b = p/q$ be the irreducible fraction representation of $b$, according to conditions of the problem $a$ and $q$ are odd. Denote by $I_n$ the set of all positive integers from interval $[qn^2, qn^2 + aq/2]$. Since $a$ and $q$ are odd, $I_n = \{qn^2 + i : i = 0, (aq - 1)/2\}$. Consider another set $P = \{(n, m) : n \in \mathbb{N}, pm \in I_n\}$. Substituting $b = p/q$ to the expression of $M(n, m)$, we get $M(n, m) = q^{-1}|qn\sqrt{n^2+a} - pm|$.
From the obvious chain of inequalities
$$n^2 < n\sqrt{n^2+a} < n^2 + a/2 \quad (*)$$
it follows that $qn^2 < qn\sqrt{n^2+a} < qn^2 + (aq)/2$. Therefore, if positive integer $mp$ doesn't belong to $I_n$, then $M(n, m) \ge q^{-1}(1 - \{aq/2\}) = 1/(2q)$ (where $\{\cdot\}$ denotes the fractional part, and since $aq$ is odd, $\{aq/2\} = 1/2$). Hence
$$\inf\{M(n, m) : (n, m) \notin P\} \ge 1/(2q), \quad (**)$$
and the zeroes of the function $M(n, m)$ are contained in $P$.
Let us find the lower bound of $M(n, m)$ for $(n, m) \in P$ for all $n$ large enough. If $(n, m) \in P$, then $pm = qn^2 + i$, where $i \in \{0, 1, \dots, (aq - 1)/2\}$. For such $n$ and $m$ we get
$$M(n, m) = q^{-1} \left| qn\sqrt{n^2+a} - qn^2 - i \right| = \frac{(q^2a - 2qi)n^2 - i^2}{q(qn\sqrt{n^2+a} + qn^2 + i)}.$$
The resulting fraction has the least numerator and the largest denominator if $i$ is maximal, i. e. $i = (aq - 1)/2$. Substituting this value of $i$ and taking into account the right inequality of $(*)$, we obtain that
$$M(n, m) \ge \frac{n^2 - (aq - 1)^2/4}{2q^2n^2 + aq^2 - q/2} \to \frac{1}{2q^2} \quad \text{as} \quad n \to \infty,$$
so, there exists an $n_0$ such that $M(n, m) > 1/(4q^2)$ for all $(n, m) \in P$ with $n \ge n_0$. In particular, because if $(**)$ holds then this means that the zeroes of $M(n, m)$ belong to the finite set $P_0 = \{(n, m) \in P : n \ge n_0\}$. Therefore $M((n, m))$ has the finite number of zeroes. Choose the minimal nonzero value of $M(n, m)$ on the set $P_0$ (say $C_0$). Then we can take $C = \min\{C_0, 1/(4q^2)\}$ as the required constant. The statement of part b) is proved.