2. Consider a billiard table whose surface is a regular 1990-sided polygon \(A_{1} A_{2} \cdots A_{1990}\). A ball is hit from a point \(P_{0}\) on \(A_{1} A_{2}\), strikes a point \(P_{1}\) on \(A_{2} A_{3}\), and subsequently hits points \(P_{2}, P_{3}, \cdots, P_{1989}, P_{1990}\) on \(A_{3} A_{4}, A_{4} A_{5}, \cdots, A_{1990} A_{1}, A_{1} A_{2}\) respectively. Let \(P_{0} A_{2} = \lambda A_{1} A_{2} (0 < \lambda < 1)\) and \(\angle P_{1} P_{0} A_{2} = \theta\).
(1) Find the range of \(\theta\);
(2) If \(P_{1990} = P_{0}\), find the value of \(\theta\).
2. (1) Let $\angle P O P_{1} A_{2}=\alpha$, then $\alpha=\pi-\frac{1990-2}{1990} \pi-\theta=\frac{\pi}{995}-\theta$.

From $\angle P_{2} P_{1} A_{3}=\angle P 0 P_{1} P_{2}=\alpha$ and $\angle P O A_{2} P_{1}=\angle P_{2} A_{3} P_{1}$, we get
$\triangle P O P_{1} A_{1} \sim \triangle P_{2} P_{1} A_{3}$,
Similarly, we have $\triangle P O P_{1} A C_{2} \backsim \triangle P_{2} P_{1} A_{3} \backsim \triangle P_{2} P_{3} A_{4} \backsim \cdots \backsim \triangle P_{1990} P_{1989} A_{1}$.
Let $P_{i} A_{i+2}=b_{i}, A_{i+2} P_{i+1}=c_{i}\left(i=0,1,2, \cdots, 1989, A_{1991}=A_{1}\right)$,
Then $b_{i+1}=a-c_{i}(i=0,1, \cdots, 1988), b_{i}=\lambda a$,
$$
\frac{\sin \alpha}{\sin \theta}=\frac{b_{0}}{c_{0}}=\frac{c_{1}}{b_{1}}=\frac{b_{2}}{c_{2}}=\cdots=\frac{b_{1988}}{c_{1988}}=\frac{c_{1989}}{b_{1989}} \text {. }
$$

Thus, $\lambda a \sin \theta=b_{0} \sin \theta=\left(a-b_{1}\right) \sin \alpha$,
$$
\begin{array}{l}
\left(a-b_{2}\right) \sin \theta=b_{1} \sin \alpha, \\
b_{2} \sin \alpha=\left(a-b_{3}\right) \sin \alpha, \\
\ldots
\end{array}
$$
$$
\begin{array}{l}
\left(a-b_{1988}\right) \sin \theta=b_{1987} \sin \alpha, \\
b_{1988} \sin \theta=\left(a-b_{1989}\right) \sin \alpha, \\
c_{1989} \sin \theta=b_{1989} \sin \alpha .
\end{array}
$$

Adding all the above equations, we get
$$
\begin{array}{l}
\left(994 a+\lambda a+c_{1989}\right) \sin \theta=995 a \sin \alpha, \\
995 a \sin \left(\frac{\pi}{995}-\theta\right)=(994+\lambda) a \sin \theta=c_{1989} \sin \theta .
\end{array}
$$

Substituting $0<c_{1989}<a$ into (1), we get
$$
0<995 a \sin \left(\frac{\pi}{995}-\theta\right)-(994+\lambda) a \sin \theta<\sin \theta,
$$

which simplifies to $0<995 \sin \frac{\pi}{995} \cos \theta-\left(994+\lambda+995 \cos \frac{\pi}{995}\right) \sin \theta<\sin \theta$,
$$
\frac{995 \sin \frac{\pi}{995}}{995\left(1+\cos \frac{\pi}{995}\right)+\lambda}<\tan \theta<\frac{995 \sin \frac{\pi}{995}}{995\left(1+\cos \frac{\pi}{995}\right)+\lambda-1} .
$$

Therefore, the range of $\theta$ is
$$
\arctan \left(\frac{995 \sin \frac{\pi}{995}}{995\left(1+\cos \frac{\pi}{995}\right)+\lambda}\right)<\theta<\arctan \left(\frac{995 \sin \frac{\pi}{995}}{995\left(1+\cos \frac{\pi}{995}\right)+\lambda-1}\right) .
$$
(2) When $p_{1990}=p_{0}$, $c_{1989}=a-b_{0}=(1-\lambda) a$, substituting into (1) we get
$$
995 a \sin \left(\frac{\pi}{995}-\theta\right)-(994+\lambda) a \sin \theta=a(1-\lambda) \sin \theta \text {, }
$$

From which we can derive
$$
\tan \theta=\frac{\sin \frac{\pi}{995}}{1+\cos \frac{\pi}{995}}=\tan \frac{\pi}{1990} \text {. }
$$

Since $\theta$ is an acute angle, we have $\theta=\frac{\pi}{1990}$.