$$\begin{aligned} \text { Given, } & \lim _{x \rightarrow 2} \frac{x^n-2^n}{x-2} =80 \\ \Rightarrow \quad & n(2)^{n-1} =80 \quad \left[\because \lim _{x \rightarrow a} \frac{x^n-a^n}{x-a}=n a^{n-1}\right]\\ \Rightarrow \quad & n(2)^{n-1} =5 \times 16 \\ \Rightarrow \quad & n \times 2^{n-1} =5 \times(2)^4 \\ \Rightarrow \quad & n \times 2^{n-1} =5 \times(2)^{5-1} \\ \therefore \quad & n =5 \end{aligned} $$

Given,

$$\begin{aligned} \lim _{x \rightarrow 0} \frac{\frac{\sin 3 x}{3 x} \cdot 3 x}{\frac{\sin 7 x}{7 x} \cdot 7 x} & =\frac{\lim _\limits{x \rightarrow 0} \frac{\sin 3 x}{3 x}}{\lim _\limits{x \rightarrow 0} \frac{\sin 7 x}{7 x}} \cdot \frac{3 x}{7 x} \\ & =\frac{3}{7} \cdot \frac{\lim _\limits{x \rightarrow 0} \frac{\sin 3 x}{3 x}}{\lim _\limits{x \rightarrow 0} \frac{\sin 7 x}{7 x}} \quad \left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right]\\ & =\frac{3}{7} \quad[\because x \rightarrow 0 \Rightarrow(k x \rightarrow 0) \text {, here } k \text { is real number }] \end{aligned}$$

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,

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