Answer: \(\mathrm{f}(x)=\frac18-2\left(x-\frac94\right)^2\); \(\mathrm{f}^{-1}\) does not exist. \(\mathrm{g}^{-1}(x)=\frac12\left(5-\mathrm{e}^{x/3}\right)\), domain \(x\leqslant3\ln5\), range \(0\leqslant\mathrm{g}^{-1}(x)\lt \frac52\).

Complete the square:

\(\mathrm{f}(x)=-2x^2+9x-10=-2\left(x-\frac94\right)^2+\frac18.\)

So

\(\mathrm{f}(x)=\frac18-2\left(x-\frac94\right)^2.\)

The turning point is \(\left(\frac94,\frac18\right)\). Since the given domain \(0\leqslant x\leqslant3\) includes this turning point, \(\mathrm{f}\) is not one-to-one on its domain. Therefore \(\mathrm{f}^{-1}\) does not exist.

For \(\mathrm{g}(x)=3\ln(5-2x)\), the \(y\)-intercept is

\(\mathrm{g}(0)=3\ln5.\)

The \(x\)-intercept is found from \(3\ln(5-2x)=0\), so \(5-2x=1\), giving \(x=2\).

The vertical asymptote is

\(x=\frac52.\)

To find the inverse, write

\(y=3\ln(5-2x).\)

Then

\(\mathrm{e}^{y/3}=5-2x.\)

Hence

\(x=\frac12\left(5-\mathrm{e}^{y/3}\right).\)

Therefore

\(\mathrm{g}^{-1}(x)=\frac12\left(5-\mathrm{e}^{x/3}\right).\)

The domain of \(\mathrm{g}^{-1}\) is the range of \(\mathrm{g}\), so

\(x\leqslant3\ln5.\)

The range of \(\mathrm{g}^{-1}\) is the domain of \(\mathrm{g}\), so

\(0\leqslant\mathrm{g}^{-1}(x)\lt \frac52.\)