Given,

$$\begin{aligned} & \lim _{x \rightarrow-3} \frac{x^3+27}{x^5+243}=\lim _{x \rightarrow-3} \frac{\frac{x^3+27}{x+3}}{\frac{x^5+243}{x+3}} \\ & =\lim _{x \rightarrow-3} \frac{\frac{x^3-(-3)^3}{x-(-3)}}{\frac{x^5-(-3)^5}{x-(-3)}}=\frac{\lim _\limits{x \rightarrow-3} \frac{x^3-(-3)^3}{x-(-3)}}{\lim _\limits{x \rightarrow-3} \frac{x^5-(-3)^5}{x-(-3)}} \quad\left[\because \lim _{x \rightarrow a} \frac{f(x)}{g(x)}=\frac{\lim _{x \rightarrow \mathrm{a}} f(x)}{\lim _{x \rightarrow a} g(x)}\right] \\ & =\frac{3(-3)^{3-1}}{5(-3)^{5-1}}=\frac{3}{5} \frac{(-3)^2}{(-3)^4} \\ & {\left[\because \lim _{x \rightarrow a} \frac{x^n-a^n}{x-a}=n a^{n-1}\right]} \\ & =\frac{3}{5(-3)^2}=\frac{3}{45}=\frac{1}{15} \end{aligned}$$

Given, $$\begin{aligned} \lim _{x \rightarrow 1 / 2}\left(\frac{8 x-3}{2 x-1}-\frac{4 x^2+1}{4 x^2-1}\right) & =\lim _{x \rightarrow 1 / 2}\left[\frac{(8 x-3)(2 x+1)-\left(4 x^2+1\right)}{\left(4 x^2-1\right)}\right] \\ & =\lim _{x \rightarrow 1 / 2}\left[\frac{16 x^2+8 x-6 x-3-4 x^2-1}{4 x^2-1}\right] \\ & =\lim _{x \rightarrow 1 / 2}\left[\frac{12 x^2+2 x-4}{4 x^2-1}\right] \\ & =\lim _{x \rightarrow 1 / 2} \frac{2\left(6 x^2+x-2\right)}{4 x^2-1} \end{aligned}$$

$$\begin{aligned} & =\lim _{x \rightarrow 1 / 2} \frac{2\left(6 x^2+4 x-3 x-2\right)}{4 x^2-1} \\ & =\lim _{x \rightarrow 1 / 2} \frac{2[2 x(3 x+2)-1(3 x+2)]}{4 x^2-1} \\ & =\lim _{x \rightarrow 1 / 2} \frac{2[(3 x+2)(2 x-1)]}{(2 x)^2-(1)^2} \\ & =\lim _{x \rightarrow 1 / 2} \frac{2(3 x+2)(2 x-1)}{(2 x-1)(2 x+1)} \\ & =\lim _{x \rightarrow 1 / 2} \frac{2(3 x+2)}{2 x-1}=\frac{2\left(3 \times \frac{1}{2}+2\right)}{2 \times \frac{1}{2}+1} \\ & =\frac{3}{2}+2=\frac{7}{2} \end{aligned}$$

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

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,

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