$\text {(I)} \begin{aligned}
3 \sin ^2 x+4 \cos ^2 x-2 & =3 \sin ^2 x+3 \cos ^2 x+\cos ^2 x-2 \\
& =3+\cos ^2 x-2=\cos ^2 x+1
\end{aligned}$
Also, $0 \leq \cos ^2 x \leq 1 \Rightarrow 1 \leq \cos ^2 x+1 \leq 2$
$\Rightarrow 1 \leq 3 \sin ^2 x+4 \cos ^2 x-2 \leq 2$
$\therefore$ Range of $\left(3 \sin ^2 x+4 \cos ^2 x-2\right)$ is $[1,2]$
$\begin{aligned}
& \text { (II) } \sin ^2 x+\cos ^4 x=\sin ^2 A+\left(1-\sin ^2 A\right)^2 \\
& =\sin ^2 A+1+\sin ^4 A-2 \sin ^2 A \\
& =-\sin ^2 A+1+\sin ^4 A \\
& =-\sin ^2 A\left(1-\sin ^2 A\right)+1 \\
& =1-\sin ^2 A \cos ^2 A \\
& =1-\left(\frac{\sin 2 A}{2}\right)^2 \\
& \because 0 \leq \sin ^2 2 x \leq 1 \Rightarrow 0 \leq \frac{\sin ^2 2 x}{4} \leq \frac{1}{4} \\
& \Rightarrow 0 \geq-\left[\frac{\sin 2 x}{2}\right]^2 \geq-\frac{1}{4} \\
& \Rightarrow 1 \geq 1-\left[\frac{\sin 2 x}{2}\right]^2 \geq 1-\frac{1}{4} \\
& \Rightarrow 1 \geq \sin ^2 A+\cos ^4 x \geq \frac{3}{4}
\end{aligned}$
$\begin{aligned}
& \text { (III) } \sin ^6 x+\cos ^6 x=\left(\sin ^2 x\right)^3+\left(\cos ^2 x\right)^3 \\
& =\left(\sin ^2 x+\cos ^2 x\right)^3-3 \sin ^2 x \cos ^2 x\left(\sin ^2 x+\cos ^2 x\right) \\
& =1-3 \sin ^2 x \cos ^2 x=1-\frac{3}{4} \sin ^2 2 x \\
& \therefore 0 \leq \sin ^2 2 x \leq 1 \\
& \Rightarrow \frac{1}{4} \leq 1-\frac{3}{4} \sin ^2 2 x \leq 1
\end{aligned}$
$\therefore$ Range of $\left(1-\frac{3}{4} \sin ^2 2 A\right)$ is $\left[\frac{1}{4}, 1\right]$.
$\begin{aligned} & \text { (IV) } \cos x \cos \left(\frac{2 \pi}{3}+x\right) \cos \left(\frac{2 \pi}{3}-x\right) \\ & =\cos x\left[\cos ^2\left(\frac{2 \pi}{3}\right) \cos ^2 x-\sin ^2\left(\frac{2 \pi}{3}\right) \sin ^2 x\right] \\ & =\cos x\left[\frac{\cos ^2 x}{4}-\frac{3 \sin ^2 x}{4}\right] \\ & =\frac{4 \cos ^3 x-3 \cos x}{4}=\frac{\cos 3 x}{4} \\ & \because-1 \leq \cos 3 x \leq 1 \\ & \Rightarrow-\frac{1}{4} \leq \frac{\cos 3 x}{4} \leq \frac{1}{4} \\ & \Rightarrow-\frac{1}{4} \leq \cos x \cos \left(\frac{2 \pi}{3}+x\right) \cos \left(\frac{2 \pi}{3}-x\right) \leq \frac{1}{4}\end{aligned}$