(i) We have, $$f(x)=\frac{3}{2-x^2}$$

$$\begin{aligned} \text{Let}\quad & y=f(x) \\ \text{Then,}\quad & y=\frac{3}{2-x^2} \Rightarrow 2-x^2=\frac{3}{y} \end{aligned}$$

$\Rightarrow \quad x^2=2-\frac{3}{y} \Rightarrow x=\sqrt{\frac{2 y-3}{y}}$

$x$ assums real values, if $2 y-3 \geq 0$ and $y>0 \Rightarrow y \geq \frac{3}{2}$

$\therefore \quad$ Range of $f=\left[\frac{3}{2}, \infty\right)$

$$\begin{aligned} \text{(ii) We know that,}\quad |x-2| & \geq 0 \Rightarrow-|x-2| \leq 0 \\ 1-|x-2| & \leq 1 \Rightarrow f(x) \leq 1 \\ \therefore \text { Range of } f & =(-\infty, 1] \end{aligned}$$

(iii) We know that, $$|x-3| \geq 0 \Rightarrow f(x) \geq 0$$

$$\therefore \quad \text { Range of } f=[0, \infty)$$

(iv) We know that,

$$\begin{aligned} & -1 \leq \cos 2 x \leq 1 \Rightarrow-3 \leq 3 \cos 2 x \leq 3 \\ & \Rightarrow \quad 1-3 \leq 1+3 \cos 2 x \leq 1+3 \Rightarrow-2 \leq 1+3 \cos 2 x \leq 1+3 \end{aligned}$$

$$\Rightarrow \quad-2 \leq f(x) \leq 4$$

$\therefore \quad$ Range of $f=[-2,4]$

$$\begin{aligned} \text{Since,}\quad |x-2| & =-(x-2), x<2 \\ x-2, x & \geq 2 \\ \text{and}\quad |2+x| & =-(2+x), x<-2 \\ (2+x), x & \geq-2 \end{aligned}$$

$\therefore \quad f(x)=|x-2|+|2+x|,-3 \leq x \leq 3$

$\begin{aligned} & =\left\{\begin{array}{cc}-(x-2)-(2+x), & -3 \leq x<-2 \\ -(x-2)+2+x, & -2 \leq x<2 \\ x-2+2+x, & 2 \leq x \leq 3\end{array}\right. \\ & =\left\{\begin{array}{cc}-2 x, & -3 \leq x<-2 \\ 4, & -2 \leq x<2 \\ 2, & 2 \leq x \leq 3\end{array}\right.\end{aligned}$

We have, $\quad f(x)=\frac{x-1}{x+1}$

(i) $f\left(\frac{1}{x}\right)=\frac{\frac{1}{x}-1}{\frac{1}{x}+1}=\frac{(1-x) / x}{(1+x) / x}=\frac{1-x}{1+x}=\frac{-(x-1)}{x+1}=-f(x)$

(ii) $f\left(-\frac{1}{x}\right)=\frac{-\frac{1}{x}-1}{-\frac{1}{x}+1}=\frac{(-1-x) / x}{(-1+x) / x} \Rightarrow f\left(-\frac{1}{x}\right)=\frac{-(x+1)}{x-1}$

$\begin{array}{ll}\text { Now, } & \frac{-1}{f(x)}=\frac{-1}{\frac{x-1}{x+1}}=\frac{-(x+1)}{x-1} \\ \therefore & f\left(-\frac{1}{x}\right)=-\frac{1}{f(x)}\end{array}$

We have, $f(x)=\sqrt{x}$ and $g(x)=x$ be two function defined in the domain $R^{+} \cup\{0\}$.

(i) $(f+g)(x)=f(x)+g(x)=\sqrt{x}+x$

(ii) $(f-g)(x)=f(x)-g(x)=\sqrt{x}-x$

(ii) $(f g)(x)=f(x) \cdot g(x)=\sqrt{x} \cdot x=x^{\frac{3}{2}}$

(iv) $\left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}=\frac{\sqrt{x}}{x}=\frac{1}{\sqrt{x}}$

We have, $\quad f(x)=\frac{1}{\sqrt{x-5}}$

$f(x)$ is defined, if $x-5>0 \Rightarrow x>5$

$\therefore \quad$ Domain of $f=(5, \infty)$

Let $$f(x)=y$$

$$\therefore \quad y=\frac{1}{\sqrt{x-5}} \Rightarrow \sqrt{x-5}=\frac{1}{y}$$

$$\begin{array}{ll} \Rightarrow & x-5=\frac{1}{y^2} \\ \therefore & x=\frac{1}{y^2}+5 \\ \because & x \in(5, \infty) \Rightarrow y \in R^{+} \end{array}$$

Hence, range of $f=R^{+}$