Solution:

Let $k = \left\lfloor \frac{1}{2x} \right\rfloor$, $k \in \mathbb{Z}$.

a. For $n=8$, the equation becomes
$$k = \left\lfloor \frac{1}{2x} \right\rfloor = 8 - 36x \Rightarrow x \neq 0 \text{ and } x = \frac{8 - k}{36}$$
Since $x \neq 0$, we have $k \neq 8$, and the last relation implies $k = \left\lfloor \frac{1}{2x} \right\rfloor = \left\lfloor \frac{18}{8 - k} \right\rfloor$. Checking signs, we see that $0 < k < 8$. By direct verification, we find the solutions $k = 3$ (hence $x = \frac{5}{36}$) and $k = 4$ (hence $x = \frac{1}{9}$).

b. From the given equation we have $x \neq 0$ and $x = \frac{2(n - k)}{n(n+1)}$. Therefore, $k \neq n$ and $k = \left\lfloor \frac{1}{2x} \right\rfloor = \left\lfloor \frac{n(n+1)}{4(n - k)} \right\rfloor$. Again, checking signs we see that $0 \leq k < n$. The last equation implies
$$\begin{gathered} k \leq \frac{n(n+1)}{4(n-k)} < k+1 \Rightarrow \left\{\begin{array}{l} (2k - n)^2 + n \geq 0 \\ (2k + 1 - n)^2 < n+1 \end{array} \Rightarrow \right. \\ \quad \Rightarrow \frac{n-1-\sqrt{n+1}}{2} < k < \frac{n-1+\sqrt{n+1}}{2} \end{gathered}$$
Conversely, if $k \in \mathbb{Z}$ satisfies (2) and $0 < k < n$, then $x = \frac{2(n - k)}{n(n+1)}$ is a solution to the given equation. It remains to note that choosing $n$ such that $\sqrt{n+1} > 2021$ ensures that there exist at least 2021 integer values of $k$ which satisfy (2).