a.) To prove the required statement it suffices to find the number $n \in M$ such that $\{\sqrt{n}\} = \min_{k \in M}\{\sqrt{k}\}$. Any number $n \in M$ can be uniquely presented as
$$n = k^2 + r, \qquad (1)$$
where $1 \le r \le 2k$, and $k = [\sqrt{n}]$ ($\cdot$ is the whole part of a number). Then
$$\{\sqrt{n}\} = \sqrt{k^2 + r} - k = \frac{r}{\sqrt{k^2 + r} + k}.$$
Since the function $y = \sqrt{x}$ is increasing on any segment $[l^2, (l+1)^2]$, $l \in \mathbb{N}$, and its values belong to $[l, l+1]$, we see that $\{\sqrt{x}\}$ has its minimal value for positive integers $x \in (l^2, (l+1)^2)$ when $x = l^2 + 1$. So the minimal value of $\{\sqrt{n}\}$ one can look for among $n$ such that $r = 1$ in representation (1), i.e., in other words, one must find the minimum of the function
$$\{\sqrt{n}\} = \frac{1}{\sqrt{k^2 + 1} + k}, \qquad (2)$$
where $n \in M$, $k = [\sqrt{n}]$. From (2) it follows that $\{\sqrt{n}\}$, $n \in M$, takes the minimal value when $k$ takes the maximal possible value. Since $44^2 = 1936 < 2015 < 2025 = 45^2$, $k$ takes its maximal value when $n = 44 \in M$. Hence for any $n \in M$ we have
$$\{\sqrt{n}\} \ge \{\sqrt{44^2 + 1}\} = \frac{1}{\sqrt{44^2 + 1} + 44} > \frac{1}{89} > 0.011.$$

b.) We have
$$\{\sqrt{44^2 + 1}\} = \frac{1}{\sqrt{44^2 + 1} + 44} < \frac{1}{88} < 0.0115.$$