Answer: \(a=\frac23\); range \(\mathbb R\); \(\mathrm f^{-1}(x)=\frac13\left(e^{x/4}+2\right)\); \(x=\frac{63}{2}\) for \(\mathrm{gg}(x)=9\).

Apply the laws of logarithms carefully, paying attention to the base of each logarithm and any restrictions on the argument.

For \(\mathrm f(x)=4\ln(3x-2)\), the logarithm is defined when \(3x-2\gt 0\), so \(x\gt \frac23\). Hence \(a=\frac23\).

As \(x\) approaches \(\frac23\) from the right, \(4\ln(3x-2)\to-\infty\). As \(x\to\infty\), \(4\ln(3x-2)\to\infty\). So the range of \(\mathrm f\) is \(\mathbb R\).

Let \(y=4\ln(3x-2)\). Then

\(\frac{y}{4}=\ln(3x-2)\), so \(e^{y/4}=3x-2\).

Thus

\(x=\frac13\left(e^{y/4}+2\right)\), and therefore

\(\mathrm f^{-1}(x)=\frac13\left(e^{x/4}+2\right)\).

The domain of \(\mathrm f^{-1}\) is \(\mathbb R\), and its range is \(x\gt \frac23\).

For the sketch, \(y=\mathrm f(x)\) has vertical asymptote \(x=\frac23\) and crosses the \(x\)-axis at \((1,0)\). The graph of \(y=\mathrm f^{-1}(x)\) is the reflection in \(y=x\), so it crosses the \(y\)-axis at \((0,1)\) and has horizontal asymptote \(y=\frac23\).

Now \(\mathrm g(x)=\sqrt{2x+1}+4\). To solve \(\mathrm{gg}(x)=9\), first solve \(\mathrm g(t)=9\):

\(\sqrt{2t+1}+4=9\), so \(\sqrt{2t+1}=5\), giving \(t=12\).

Therefore \(\mathrm g(x)=12\):

\(\sqrt{2x+1}+4=12\), so \(\sqrt{2x+1}=8\).

Hence \(2x+1=64\), and \(x=\frac{63}{2}\).

The result has been obtained using the exact condition from the question, so no additional solutions are introduced.