(I) $\cos x+\sin x=1\Rightarrow \sqrt{2}\sin \left(x+\frac{\pi }{4}\right)=1\Rightarrow \sin \left(x+\frac{\pi }{4}\right)=\frac{1}{\sqrt{2}}$. Solutions: $x+\frac{\pi }{4}=\frac{\pi }{4}+2n\pi$ or $x+\frac{\pi }{4}=\frac{3\pi }{4}+2n\pi$, i.e., $x=2n\pi$ or $x=\frac{\pi }{2}+2n\pi$. In $\left[-\frac{2\pi }{3},\frac{2\pi }{3}\right]$, valid values are $x=0,\frac{\pi }{2}$. Two elements $\to$ (P).

(II) $\sqrt{3}\tan 3x=1\Rightarrow \tan 3x=\frac{1}{\sqrt{3}}\Rightarrow 3x=\frac{\pi }{6}+n\pi \Rightarrow x=\frac{\pi }{18}+\frac{n\pi }{3}$. In $\left[-\frac{5\pi }{18},\frac{5\pi }{18}\right]$, valid values are $x=\frac{\pi }{18}$ (for $n=0$) and $x=-\frac{5\pi }{18}$ (for $n=-1$). Two elements $\to$ (P).

(III) $2\cos 2x=\sqrt{3}\Rightarrow \cos 2x=\frac{\sqrt{3}}{2}\Rightarrow 2x=\pm \frac{\pi }{6}+2n\pi \Rightarrow x=\pm \frac{\pi }{12}+n\pi$. In $\left[-\frac{6\pi }{5},\frac{6\pi }{5}\right]$, the valid integers $n$ for each branch yield six total values: $x=\pm \frac{\pi }{12},\pm \frac{\pi }{12}\pm \pi$ (selecting those in range). Six elements $\to$ (T).

(IV) $\sin x-\cos x=1\Rightarrow \sqrt{2}\sin \left(x-\frac{\pi }{4}\right)=1\Rightarrow x-\frac{\pi }{4}=\frac{\pi }{4}+2n\pi$ or $\frac{3\pi }{4}+2n\pi$, i.e., $x=\frac{\pi }{2}+2n\pi$ or $x=\pi +2n\pi$. In $\left[-\frac{7\pi }{4},\frac{7\pi }{4}\right]$, valid values: $x=\frac{\pi }{2},\pi ,-\pi ,-\frac{3\pi }{2}$. Four elements $\to$ (R).

Matching gives (I) $\to$ (P); (II) $\to$ (P); (III) $\to$ (T); (IV) $\to$ (R), which is option (B).