One, (25 points) Find all values of $\alpha$, $\alpha \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, such that the system of equations
$$
\left\{\begin{array}{l}
y=\operatorname{arasin}\left(\sin \frac{1}{x}\right), \\
y=\operatorname{tg} \alpha\left(x-\frac{1}{10 \pi}\right)
\end{array}\right.
$$

has exactly 10 solutions for $x \geqslant \frac{1}{10 \pi}$.
$-, y=\arcsin \left(\sin \frac{1}{x}\right)$
$$
=\left\{\begin{array}{c}
\frac{1}{x}, x \geqslant \frac{2}{\pi} \\
\pi-\frac{1}{x}, \frac{2}{3 \pi} \leqslant x < \frac{2}{\pi} \\
-\pi-\frac{1}{x}, x < \frac{2}{3 \pi}
\end{array}\right.
$$

When $x > 0$, the above curve intersects with the line at most 9 points in $\left[\frac{1}{10 \pi}, \frac{1}{2 \pi}\right]$, so the required $u$ is
$$
\left\{\begin{array}{l}
y=\arcsin \left(\sin \frac{1}{x}\right), \\
y=\operatorname{tg} \alpha\left(x-\frac{1}{10 \pi}\right) .
\end{array}\right.
$$

Since $10-3 \sqrt{10}>\frac{1}{2}$, we know
$$
\frac{1}{(10-3 \sqrt{10}) \pi} \in\left(\frac{2}{3 \pi}, \frac{2}{\pi}\right),
$$

thus $\alpha=\operatorname{arctg}\left[(190-60 \sqrt{10}) \pi^{2}\right]$.
For $x \in\left(-\frac{\pi}{2}, 0\right)$, the semi-straight line passing through $\left(\frac{1}{10 \pi}, 0\right)$ and $\left(\frac{2}{3 \pi},-\frac{\pi}{2}\right)$ intersects the curve at 9 points, at this time the slope is
$$
\operatorname{tg} \alpha=\frac{-\frac{\pi}{2}}{\frac{2}{3 \pi}-\frac{1}{10 \pi}}=-\frac{15}{17} \pi^{2},
$$

thus, when $\alpha$ satisfies
$$
\operatorname{arctg}\left(-\frac{15}{17} \pi^{2}\right)<\alpha \leqslant 0
$$

the semi-straight line intersects the curve at 10 points.
Therefore, the solution set of $a$ is
$$
\begin{array}{l}
\left\{\operatorname{arctg}\left[(190-60 \sqrt{10}) \pi^{2}\right]\right\} \\
U\left(\operatorname{arctg}\left(-\frac{15}{17} \pi^{2}\right), 0\right] .
\end{array}
$$