Answer: \(x\gt \frac12\); \(\mathrm{f}^{-1}(x)=\frac12(1+e^{x/4})\), domain \(x\in\mathbb{R}\); \(x=\frac72\).

(a)(i) For \(\ln(2x-1)\) to be defined,

\(2x-1\gt 0.\)

Therefore

\(x\gt \frac12.\)

(a)(ii) Let

\(y=4\ln(2x-1).\)

Then

\(\frac{y}{4}=\ln(2x-1),\)

so

\(e^{y/4}=2x-1.\)

Hence

\(x=\frac12(1+e^{y/4}).\)

Therefore

\(\mathrm{f}^{-1}(x)=\frac12(1+e^{x/4}).\)

The domain of \(\mathrm{f}^{-1}\) is the range of \(\mathrm{f}\), which is all real numbers, so

\(x\in\mathbb{R}.\)

(b) Since \(\mathrm{gh}(x)=\mathrm{g}(\mathrm{h}(x))\),

\(\mathrm{gh}(x)=\sqrt{2x-3}+5.\)

Solve

\(\sqrt{2x-3}+5=7.\)

Then

\(\sqrt{2x-3}=2.\)

Squaring gives

\(2x-3=4,\)

and hence

\(x=\frac72.\)