Let a function $f:\left(2,\mathrm{\infty }\right)\to \left[0,\mathrm{\infty }\right)$ defined as $f\left(x\right)=\left|\frac{x-3}{x-2}\right|$, then $f$ is

$f\left(x\right)=\left|1-\frac{1}{\left(x-2\right)}\right|$
$\because 2<x<\mathrm{\infty }$
$0<x-2<\mathrm{\infty }\Rightarrow \mathrm{\infty }>\frac{1}{x-2}>0$
$-\mathrm{\infty }<-\frac{1}{\left(x-2\right)}<0\Rightarrow -\mathrm{\infty }<1-\frac{1}{\left(x-2\right)}<1$
$\because 0\le \left|1-\frac{1}{\left(x-2\right)}\right|<\mathrm{\infty }$
$\therefore$ Range of $f\left(x\right)\in \left[0,\mathrm{\infty }\right)=$ co-domain

Hence, $f\left(x\right)$ is surjective
Let, $f\left(x\right)=\frac{1}{2}\Rightarrow \left|1-\frac{1}{x-2}\right|=\frac{1}{2}$
$1-\frac{1}{\left(x-2\right)}=\frac{1}{2}, 1-\frac{1}{\left(x-2\right)}=-\frac{1}{2}$
$\frac{1}{x-2}=\frac{1}{2}, \frac{1}{x-2}=\frac{3}{2}$
$x=4, x=\frac{8}{3}$
$\therefore f\left(4\right)=f\left(\frac{8}{3}\right)=\frac{1}{2}$
$\therefore f\left(x\right)$ is many-one (not injective)