One. (20 points) Let the function
$$
f(x)=\cos x \cdot \cos (x-\theta)-\frac{1}{2} \cos \theta
$$

where, $x \in \mathbf{R}, 0<\theta<\pi$. It is known that when $x=\frac{\pi}{3}$, $f(x)$ achieves its maximum value.
(1) Find the value of $\theta$;
(2) Let $g(x)=2 f\left(\frac{3}{2} x\right)$, find the minimum value of the function $g(x)$ on $\left[0, \frac{\pi}{3}\right]$.
(1) Notice
$$
\begin{array}{l}
f(x)=\cos x(\cos x \cdot \cos \theta+\sin x \cdot \sin \theta)-\frac{1}{2} \cos \theta \\
=\frac{1+\cos 2 x}{2} \cos \theta+\frac{1}{2} \sin 2 x \cdot \sin \theta-\frac{1}{2} \cos \theta \\
=\frac{1}{2} \cos (2 x-\theta) .
\end{array}
$$

By $f(x)_{\max }=f\left(\frac{\pi}{3}\right)=\frac{1}{2} \cos \left(\frac{2 \pi}{3}-\theta\right)=\frac{1}{2}$, we get $\cos \left(\frac{2 \pi}{3}-\theta\right)=1$.
Since $0<\theta<\pi$, thus, $\theta=\frac{2 \pi}{3}$.
(2) From (1), we know $f(x)=\frac{1}{2} \cos \left(2 x-\frac{2 \pi}{3}\right)$. Then $g(x)=2 f\left(\frac{3}{2} x\right)=\cos \left(3 x-\frac{2 \pi}{3}\right)$.
Since $0 \leqslant x \leqslant \frac{\pi}{3}$, therefore,
$$
-\frac{2 \pi}{3} \leqslant 3 x-\frac{2 \pi}{3} \leqslant \frac{\pi}{3} \text {. }
$$

Hence, when $3 x-\frac{2 \pi}{3}=-\frac{2 \pi}{3}$, i.e., $x=0$,
$$
g(x)_{\text {min }}=-\frac{1}{2} .
$$