(a)(i) $(a, 0) \rightarrow\left(\frac{a+3}{2}, 0\right) \quad(0, b) \rightarrow\left(\frac{3}{2}, b\right) \left(\frac{a+3}{2}, 0\right)$ is the only point. Asymptote: $y=k$
(ii) $(a, 0) \rightarrow(0, a),  (0, b) \rightarrow(b, 0)$ Asymptote: $x=k$

(b)(i) $ \mathrm{g}^{-1}(x)=\ln \left(\frac{\mathrm{e}^x+5}{\mathrm{e}^x-1}\right),  \mathrm{D}_{\mathrm{g}^{-1}}=\mathrm{R}_{\mathrm{g}}=(0, \infty)$ (ii) $x=2 \pm \sqrt{(2+\ln 7)(4-\ln 7)}$

(a)(i) $(a, 0) \rightarrow\left(\frac{a+3}{2}, 0\right) \quad(0, b) \rightarrow\left(\frac{3}{2}, b\right)$

$\left(\frac{a+3}{2}, 0\right)$ is the only point. Asymptote: $y=k$
(ii) $(a, 0) \rightarrow(0, a),  (0, b) \rightarrow(b, 0)$
Asymptote: $x=k$

(b)(i) $y=\ln \left(\frac{\mathrm{e}^x+5}{\mathrm{e}^x-1}\right) \Rightarrow \mathrm{e}^y=\frac{\mathrm{e}^x+5}{\mathrm{e}^x-1} \Rightarrow\left(\mathrm{e}^y-1\right) \mathrm{e}^x=\mathrm{e}^y+5$
$\Rightarrow \mathrm{e}^x=\frac{\mathrm{e}^y+5}{\mathrm{e}^y-1} \Rightarrow x=\ln \left(\frac{\mathrm{e}^y+5}{\mathrm{e}^y-1}\right)$
$\therefore \mathrm{g}^{-1}(x)=\ln \left(\frac{\mathrm{e}^x+5}{\mathrm{e}^x-1}\right)$
$\mathrm{D}_{\mathrm{g}^{-1}}=\mathrm{R}_{\mathrm{g}}=(0, \infty)$

(ii) Note: $\mathrm{g}$ is a self-inverse function
$\mathrm{g}^{2021} \mathrm{~h}(x)=\ln 2$
$\operatorname{gh}(x)=\ln 2$

$h(x)=g^{-1}(\ln 2)=\ln 7$
$1+\sqrt{9-(x-2)^2}=\ln 7$

$9-(x-2)^2=(\ln 7-1)^2$
$(x-2)^2=9-(\ln 7-1)^2=(2+\ln 7)(4-\ln 7)$
$x=2 \pm \sqrt{(2+\ln 7)(4-\ln 7)}$