Answer: (i) \(x=\frac13\). (ii) \(f^{-1}(x)=\frac{e^x-2}{3}\). (iii) \(x=\ln2\).

(i) Since

\(f(x)=\ln(3x+2),\)

we have

\(g(f(x))=e^{2\ln(3x+2)}-4.\)

Using \(e^{2\ln u}=u^2\),

\(g(f(x))=(3x+2)^2-4.\)

So \(gf(x)=5\) gives

\((3x+2)^2-4=5.\)

Hence

\((3x+2)^2=9.\)

Since \(3x+2\gt0\),

\(3x+2=3.\)

Therefore

\(x=\frac13.\)

(ii) Let \(y=\ln(3x+2)\). Then

\(e^y=3x+2.\)

So

\(x=\frac{e^y-2}{3}.\)

Hence

\(f^{-1}(x)=\frac{e^x-2}{3}.\)

(iii) Solve

\(\frac{e^x-2}{3}=e^{2x}-4.\)

Multiplying by \(3\),

\(e^x-2=3e^{2x}-12.\)

So

\(3e^{2x}-e^x-10=0.\)

Let \(u=e^x\). Then

\(3u^2-u-10=0.\)

Factorising,

\((3u+5)(u-2)=0.\)

Since \(u=e^x\gt0\), \(u=2\). Therefore

\(\boxed{x=\ln2}.\)