Answer: $8 \pi$.
Solution: Let the projection of point $P$ on plane $\alpha$ be $O$. From the given conditions, $\frac{O P}{O Q}=\tan \angle O Q P \in\left[\frac{\sqrt{3}}{3}, \sqrt{3}\right]$, i.e., $O Q \in[1,3]$. Therefore, the area of the required region is $\pi \cdot 3^{2}-\pi \cdot 1^{2}=8 \pi$.
2. Let the distance from point $P$ to plane $\alpha$ be $\sqrt{3}$, and point $Q$ lies on plane $\alpha$ such that the angle between line $P Q$ and $\alpha$ is no less than $30^{\circ}$ and no more than $60^{\circ}$. Then the area of the region formed by such points $Q$ is $\qquad$.