a.
As $ma$, $na$ are natural numbers, the equality implies $\{m\sqrt{a}\} = \{-n\sqrt{a}\}$.
Two numbers have the same fractional part if and only if their difference is an integer, whence $(m+n)\sqrt{a} \in \mathbb{Z}$, which is absurd.

b.
Again, let us suppose that there is a natural number $N$, for which there exist two not equal numbers $m$, $n$ which are different from zero, such that $N = [m(a + \sqrt{a})] = [n(a - \sqrt{a})]$. Then $N \leq m(a + \sqrt{a}) < N + 1$ and $N \leq n(a - \sqrt{a}) < N + 1$; moreover, the inequalities are strict, because the terms in the middle are irrational numbers.
We rewrite the inequalities as
$$\frac{N}{a + \sqrt{a}} < m < \frac{N + 1}{a + \sqrt{a}}, \quad \frac{N}{a - \sqrt{a}} < n < \frac{N + 1}{a - \sqrt{a}},$$
and thus, by addition, we get $N \frac{2}{a-1} < m + n < (N+1) \frac{2}{a-1}$.
From here, $N < \frac{a-1}{2}(m+n) < N+1$, which is a contradiction, because the term in the middle is natural ($a$ is odd).