Given,

$$\begin{aligned} & \lim _{x \rightarrow 0} \frac{\sin ^2 2 x}{\sin ^2 4 x}=\lim _{x \rightarrow 0} \frac{\sin ^2 2 x}{[\sin 2(2 x)]^2} \\ & =\lim _{x \rightarrow 0} \frac{\sin ^2 2 x}{(2 \sin 2 x \cos 2 x)^2} \\ & =\lim _{x \rightarrow 0} \frac{\sin ^2 2 x}{4 \sin ^2 2 x \cos ^2 2 x} \quad[\because \sin 2 \theta=2 \sin \theta \cos \theta] \\ & =\lim _{x \rightarrow 0} \frac{1}{4 \cos ^2 2 x}=\frac{1}{4} \quad {[\because \cos 0=1]} \end{aligned}$$

Given,

$$\begin{aligned} \lim _{x \rightarrow 0} \frac{1-\cos 2 x}{x^2} & =\lim _{x \rightarrow 0} \frac{1-1+2 \sin ^2 x}{x^2} \quad \left[\because \cos 2 x=1-2 \sin ^2 x\right]\\ & =\lim _{x \rightarrow 0} \frac{2 \sin ^2 x}{x^2}=2 \lim _{x \rightarrow 0} \frac{\sin ^2 x}{x^2} \\ & =2 \lim _{x \rightarrow 0}\left(\frac{\sin x}{x}\right)^2 \quad\left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right]\\ & =2 \times 1=2 \end{aligned}$$

Given,

$$\begin{aligned} \lim _{x \rightarrow 0} \frac{2 \sin x-\sin 2 x}{x^3} & =\lim _{x \rightarrow 0} \frac{2 \sin x-2 \sin x \cos x}{x^3} \quad [\because \sin2x=2\sin x \cos x]\\ & =\lim _{x \rightarrow 0} \frac{2 \sin x(1-\cos x)}{x^3} \\ & =2 \lim _{x \rightarrow 0} \frac{\sin x}{x} \cdot \lim _{x \rightarrow 0}\left(\frac{1-\cos x}{x^2}\right) \\ & =2 \cdot 1 \lim _{x \rightarrow 0} \frac{1-\cos x}{x^2} \quad \left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right]\\ & =2 \lim _{x \rightarrow 0} \frac{1-1+2 \sin ^2 \frac{x}{2}}{x^2}=2 \lim _{x \rightarrow 0} \frac{2 \sin ^2 \frac{x}{2}}{4 \times \frac{x^2}{4}} \\ & =\frac{2 \cdot 2}{4} \lim _{x \rightarrow 0}\left(\frac{\sin \frac{x}{2}}{\frac{x}{2}}\right)^2=\lim _{x \rightarrow 0}\left(\frac{\sin \frac{x}{2}}{\frac{x}{2}}\right)^2=1 \end{aligned}$$

Given, $\lim _\limits{x \rightarrow 0} \frac{1-\cos m x}{1-\cos n x}=\lim _\limits{x \rightarrow 0} \frac{1-1+2 \sin ^2 \frac{m x}{2}}{1-1+2 \sin ^2 \frac{n x}{2}}$ $\left[\begin{array}{l}\because \quad \cos m x=1-2 \sin ^2 \frac{m x}{2} \\ \text { and } \sin n x=1-2 \sin ^2 \frac{n x}{2}\end{array}\right]$

$$=\lim _\limits{x \rightarrow 0} \frac{\sin ^2 \frac{m x}{2}}{\sin ^2 \frac{n x}{2}}=\lim _\limits{x \rightarrow 0} \frac{\frac{\sin ^2 \frac{m x}{2}}{\left(\frac{m x}{2}\right)^2} \cdot\left(\frac{m x}{2}\right)^2}{\frac{\sin ^2 \frac{n x}{2}}{\left(\frac{n x}{2}\right)^2} \cdot\left(\frac{n x}{2}\right)^2}=\frac{\lim _\limits{x \rightarrow 0}\left(\frac{\sin \frac{m x}{2}}{\frac{m x}{2}}\right)^2}{\lim _\limits{x \rightarrow 0}\left(\frac{\sin \frac{n x}{2}}{\frac{n x}{2}}\right)^2} \cdot \frac{m^2 \frac{x^2}{4}}{n^2 \frac{x^2}{4}}$$

$$=\frac{m^2}{n^2} \cdot \frac{\lim _\limits{x \rightarrow 0}\left(\frac{\sin \frac{m x}{2}}{\frac{m x}{2}}\right)^2}{\lim _\limits{x \rightarrow 0}\left(\frac{\sin \frac{n x}{2}}{\frac{n x}{2}}\right)^2}=\frac{m^2}{n^2} \quad\left[\because \lim _\limits{x \rightarrow 0} \frac{\sin x}{x}=1\right]$$

$[\because x\to0\Rightarrow k ~x\to 0]$

Given, $\lim _\limits{x \rightarrow \pi / 3} \frac{\sqrt{1-\cos 6 x}}{\sqrt{2}\left(\frac{\pi}{3}-x\right)}=\lim _\limits{x \rightarrow \pi / 3} \frac{\sqrt{1-1+2 \sin ^2 3 x}}{\sqrt{2}\left(\frac{\pi}{3}-x\right)}\quad \left[\because \cos 2 x=1-2 \sin ^2 x\right]$

$$\begin{array}{ll} =\lim _\limits{x \rightarrow \pi / 3} \frac{\sqrt{2} \sin 3 x}{\sqrt{2}\left(\frac{\pi}{3}-x\right)}=\lim _\limits{x \rightarrow \pi / 3} \frac{\sin 3 x}{\frac{\pi}{3}-x} & \\ =\lim _\limits{x \rightarrow \pi / 3} \frac{\sin (\pi-3 x)}{\frac{\pi-3 x}{3}} & {[\because \sin (\pi-\theta)=\sin \theta]} \\ =3 \lim _\limits{x \rightarrow \pi / 3} \frac{\sin (\pi-3 x)}{(\pi-3 x)}=3 \times 1 & {\left[\because \lim _\limits{x \rightarrow 0} \frac{\sin x}{x}=1\right]} \\ =3 & {\left[\because x \rightarrow \frac{\pi}{3} \Rightarrow\left(x-\frac{\pi}{3}\right) \rightarrow 0\right]} \end{array}$$