Five. (20 points) Given the function
$$
y=f(x)=\cos m x \cdot \cos ^{m} x-\sin m x \cdot \sin ^{m} x-\cos ^{m} 2 x
$$

whose value is independent of the real number $x$. Draw tangents $P A$ and $P B$ from any point $P$ on its graph to the circle $x^{2}+(y-2)^{2}=1$, with the points of tangency being $A$ and $B$.
(1) Find the value of the positive integer $m$;
(2) Find the minimum value of the area of $\triangle P A B$.
(1) Let $x=0$, for all $m$. Then $f(0)=0$. Since the value of $y=f(x)$ is independent of $x$, we have $f\left(\frac{\pi}{m}\right)=f(0)=0$, which gives
$$
-\cos ^{m} \frac{\pi}{m}=\cos ^{m} \frac{2 \pi}{m} \text {. }
$$

From the above equation and $m \in \mathbf{N}_{+}$, it is easy to see that $m$ is a positive odd number, hence
$$
-\cos \frac{\pi}{m}=\cos \frac{2 \pi}{m} \text {. }
$$

Therefore, $\frac{2 \pi}{m}=2 k \pi+\pi \pm \frac{\pi}{m}(k \in \mathbf{Z})$.
Thus, $m=1$ or 3.
Substituting $m=1$ or $m=3$ into $f(x)=0$ for verification, we find that $m=1$.
(2) From (1), we know that the original function is the constant function $y=0$. Let $P(a, 0)$, it is easy to see that the center of the circle $x^{2}+(y-2)^{2}=1$ is $Q(0,2)$, with radius $r=1, A Q=B Q=1, P Q=\sqrt{a^{2}+4}$.

Also, $A Q \perp P A, B Q \perp P B$ (as shown in Figure 5), so,
$$
\begin{array}{l}
A P=\sqrt{P Q^{2}-A Q^{2}} \\
=\sqrt{a^{2}+3}=B P .
\end{array}
$$

Connecting $A B$ and intersecting $P Q$ at point $H$, we have $A H \perp P Q$.
Thus,
$$
\begin{array}{l}
A H=\frac{A P \cdot A Q}{P Q}=\frac{\sqrt{a^{2}+3}}{\sqrt{a^{2}+4}}=B H, \\
P H=\sqrt{A P^{2}-A H^{2}}=\frac{a^{2}+3}{\sqrt{a^{2}+4}} .
\end{array}
$$

Therefore, $S_{\triangle P A B}=\frac{1}{2} A B \cdot P H=\frac{\left(a^{2}+3\right) \sqrt{a^{2}+3}}{a^{2}+4}$
$$
=\sqrt{a^{2}+3}\left(1-\frac{1}{a^{2}+4}\right) \text {. }
$$

When $a=0$, $S_{\triangle P R B}$ reaches its minimum value $\frac{3 \sqrt{3}}{4}$.