$$
\begin{array}{c}
\text { Five. (20 points) Given } \theta_{1}, \theta_{2}, \cdots, \theta_{n} \in \mathbf{R} \text {, and } \\
f=\sin \left(\theta_{1}-\theta_{2}\right)+\sin \left(\theta_{2}-\theta_{3}\right)+\cdots+\sin \left(\theta_{n}-\theta_{1}\right),
\end{array}
$$

Here $n \geqslant 3$.
(1) When $\left\{\theta_{k}\right\}$ is monotonic and $\left|\theta_{n}-\theta_{1}\right| \leqslant 2 \pi$, find the range of $f$;
(2) When $\theta_{k}$ are arbitrary real numbers $(k=1,2, \cdots, n)$, find the maximum and minimum values of $f$ in this case.
(1) Suppose $|\theta_{k}|$ is decreasing, then $\theta_{i}-\theta_{i+1} \in [0, 2\pi)$.
For $z_{k} = \mathrm{e}^{\mathrm{i} \theta_{k}}$, we have
$\sin \left(\theta_{k}-\theta_{k+1}\right) = \operatorname{Im}\left(z_{k} \overline{z_{k+1}}\right)$.
Thus, $f = \operatorname{Im}\left(\sum_{k=1}^{n} z_{k} \overline{z_{k+1}}\right)$, where $z_{n+1} = z_{1}$.
Given that $z_{1}, z_{2}, \cdots, z_{n}$ are $n$ points arranged sequentially on the unit circle and do not exceed one circumference, $f = 2 S_{n}$, where $S_{n}$ is the area of the convex $n$-sided polygon formed by these $n$ points.
When these $n$ points form a regular $n$-sided polygon, $f$ is maximized, and
$f_{\text{max}} = n \sin \frac{2 \pi}{n}$.
By symmetry, $f_{\min} = -n \sin \frac{2 \pi}{n}$.
Since the area of the convex $n$-sided polygon can vary from 0 to $n \sin \frac{2 \pi}{n}$, the range of $f$ is $\left[-n \sin \frac{2 \pi}{n}, n \sin \frac{2 \pi}{n}\right]$.
(2) Let $\theta_{k}-\theta_{k+1} = x_{k} - 2 m_{k} \pi$, where $m_{k} \in \mathbf{Z}$ and $x_{k} \in [0, \pi]$.
If $x_{i} \in (\pi, 2 \pi)$, then
$$
f \leqslant n-1 < f_{\text{max}} = n \sin \frac{2\left[\frac{n+1}{4}\right] \pi}{n},
$$

where $\left[\frac{n+1}{4}\right]$ denotes the greatest integer not exceeding $\frac{n+1}{4}$.
Therefore, $x_{1} + x_{2} + \cdots + x_{n} = 2 m \pi$, where $m = m_{1} + m_{2} + \cdots + m_{n} \in \mathbf{Z}$, and $0 \leqslant m \leqslant \frac{n}{2}$.
By Jensen's inequality, we have
$$
f = \sum_{k=1}^{n} \sin x_{k} \leqslant n \sin \left(\frac{x_{1} + x_{2} + \cdots + x_{n}}{n}\right) = \sin \frac{2 m \pi}{n}.
$$

When $4 \mid n$, take $m = \frac{n}{4}$.
Let $\theta_{k} - \theta_{k+1} = \frac{\pi}{2}$, $\theta_{1} = 0$, then $f_{\text{max}} = n$;
Let $\theta_{k} - \theta_{k+1} = -\frac{\pi}{2}$, $\theta_{1} = 0$, then $f_{\text{min}} = -n$.
When $4 \nmid n$, by the monotonicity of $\sin x$ in $(0, \pi)$ and the convexity of integers, we get
$$
\begin{array}{l}
f_{\text{max}} = n \sin \frac{2\left[\frac{n+1}{4}\right] \pi}{n}, f_{\text{min}} = -n \sin \frac{2\left[\frac{n+1}{4}\right] \pi}{n}. \\
\text{Let } \theta_{1} = 0, \theta_{k+1} = \mp \frac{2 k\left[\frac{n+1}{4}\right] \pi}{n} \text{ to achieve the above.}
\end{array}
$$

Thus,
$$
f_{\text{max}} = n \sin \frac{2\left[\frac{n+1}{4}\right] \pi}{n}, f_{\text{min}} = -n \sin \frac{2\left[\frac{n+1}{4}\right] \pi}{n}.
$$