(a) Let $r$ be the unique positive integer for which $r^2 \leq n < (r+1)^2$. Write $n = r^2 + s$. Then we have $0 \leq s \leq 2r$. We discuss in two cases according to the parity of $s$.

- Case 1. $s$ is even.
Consider the number $\left(r + \frac{s}{2r}\right)^2 = r^2 + s + \left(\frac{s}{2r}\right)^2$. We find that
$$n = r^2 + s \leq r^2 + s + \left(\frac{s}{2r}\right)^2 \leq r^2 + s + 1 = n + 1$$
It follows that
$$\sqrt{n} \leq r + \frac{s}{2r} \leq \sqrt{n+1}$$
Since $s$ is even, we can choose the fraction $r + \frac{s}{2r} = \frac{r^2 + (s/2)}{r}$ since $r \leq \sqrt{n}$.

- Case 2. $s$ is odd.
Consider the number $\left(r + 1 - \frac{2r + 1 - s}{2(r + 1)}\right)^2 = (r + 1)^2 - (2r + 1 - s) + \left(\frac{2r + 1 - s}{2(r + 1)}\right)^2$. We find that
$$\begin{aligned} n = r^2 + s = (r + 1)^2 - (2r + 1 - s) & \leq (r + 1)^2 - (2r + 1 - s) + \left(\frac{2r + 1 - s}{2(r + 1)}\right)^2 \\ & \leq (r + 1)^2 - (2r + 1 - s) + 1 = n + 1 \end{aligned}$$
It follows that
$$\sqrt{n} \leq r + 1 - \frac{2r + 1 - s}{2(r + 1)} \leq \sqrt{n+1}$$
Since $s$ is odd, we can choose the fraction $(r + 1) - \frac{2r + 1 - s}{2(r + 1)} = \frac{(r + 1)^2 - r + ((s - 1)/2)}{r + 1}$ since $r + 1 \leq \sqrt{n} + 1$.

(b) We show that for every positive integer $r$, there is no fraction $\frac{a}{b}$ with $b \leq \sqrt{r^2 + 1}$ such that $\sqrt{r^2 + 1} \leq \frac{a}{b} \leq \sqrt{r^2 + 2}$. Suppose on the contrary that such a fraction exists. Since $b \leq \sqrt{r^2 + 1} < r + 1$ and $b$ is an integer, we have $b \leq r$. Hence,
$$(b r)^2 < b^2 (r^2 + 1) \leq a^2 \leq b^2 (r^2 + 2) \leq b^2 r^2 + 2br < (br + 1)^2$$
This shows the square number $a^2$ is strictly bounded between the two consecutive squares $(br)^2$ and $(br + 1)^2$, which is impossible. Hence, we have found infinitely many $n = r^2 + 1$ for which there is no fraction of the desired form.