Answer: \(a=\frac43\), range \(\mathbb{R}\), and \(\mathrm{f}^{-1}(x)=\frac{4+\mathrm{e}^{x/2}}{3}\).

The logarithm requires

\(3x-4\gt 0,\)

so

\(x\gt \frac43.\)

Therefore the least possible value is

\(a=\frac43.\)

As \(x\to\frac43^+\), \(2\ln(3x-4)\to-\infty\). As \(x\to\infty\), \(2\ln(3x-4)\to\infty\). Hence the range is

\(\mathbb{R}.\)

To find the inverse, let

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

Then

\(\mathrm{e}^{y/2}=3x-4.\)

So

\(x=\frac{4+\mathrm{e}^{y/2}}{3}.\)

Therefore

\(\mathrm{f}^{-1}(x)=\frac{4+\mathrm{e}^{x/2}}{3}.\)

For \(y=\mathrm{f}(x)\), the vertical asymptote is \(x=\frac43\), and the \(x\)-intercept is found from

\(2\ln(3x-4)=0,\)

so \(3x-4=1\) and \(x=\frac53\). There is no \(y\)-intercept because the domain is \(x\gt \frac43\).

For \(y=\mathrm{f}^{-1}(x)\), the horizontal asymptote is \(y=\frac43\). The \(y\)-intercept is

\(\frac{4+\mathrm{e}^{0}}{3}=\frac53,\)

and there is no \(x\)-intercept.

The two graphs are reflections in the line \(y=x\), so they should be sketched symmetrically about that line.