17. With $A$ as the center and $2 \cos \theta\left(\frac{\pi}{4}<\theta<\frac{\pi}{2}\right)$ as the radius, there is a point $B$ outside the circle; it is known that $|A B|=2 \sin \theta$, and the center of the circle passing through point $B$ and externally tangent to $\odot A$ at point $T$ is $M$.
(1) When $\theta$ takes a certain value, explain what the curve of the trajectory $P$ of point $M$ is.
(2) Point $M$ is a moving point on the trajectory $P$, and point $N$ is a moving point on $\odot A$. Denote the minimum value of $|M N|$ as $f(\theta)$ (proof not required), and find the range of $f(\theta)$.
(3) If the range of $\theta$ in the problem statement is changed to $\left(0<\theta<\frac{\pi}{4}\right)$, and the position of point $B$ is changed to inside $\odot A$, with all other conditions remaining the same, the trajectory of point $M$ is denoted as $P$. Propose a meaningful question with the same structure as (2) (no solution required).
17. ( I ) Connect $M T$, $M A$, and $M B$. Clearly, points $M$, $T$, and $A$ are collinear, and $|M A| - |M T| = |A T| = 2 \cos \theta$.

Also, $|M T| = |M B|$, so $|M A| - |M B| = 2 \cos \theta$, where $A B = 2 \sin \theta$.
Thus, the locus of point $M$ is the branch of a hyperbola with foci at $A$ and $B$, and the real axis length is $2 \cos \theta$, which is closer to focus $B$.
(II) Clearly, the minimum value of $|M N|$, $f(\theta)$, is the $|M T|$ shown in the figure (where $N$ and $T$ coincide).
$$
|M T| = |M A| - |A T| = \sin \theta + \cos \theta - 2 \cos \theta = \sin \theta - \cos \theta
$$
Thus, $f(\theta) = \sin \theta - \cos \theta = \sqrt{2} \sin \left(\theta - \frac{\pi}{4}\right)$.
Given $\frac{\pi}{4} < \theta < \frac{\pi}{2}$, we have $0 < \theta - \frac{\pi}{4} < \frac{\pi}{4}$, which implies
$$
0 < f(\theta) < 1 \text{. }
$$
(III) Let point $M$ be a moving point on the locus $P$, and point $N$ be a moving point on $\odot A$. Denote the maximum value of $|M N|$ as $g(\theta)$, and find the range of $g(\theta)$.