Answer: \(\mathrm{g}^{-1}(x)=\ln\left(\frac{3}{2-x}-2\right)\), domain \(1\leqslant x\lt 2\).

For \(\mathrm{f}(x)=2\mathrm{e}^{-x}+3\), the \(y\)-intercept is

\(\mathrm{f}(0)=5,\)

and the horizontal asymptote is \(y=3\). Since \(2\mathrm{e}^{-x}\gt 0\), the graph does not meet the \(x\)-axis.

The graph of \(y=\mathrm{f}^{-1}(x)\) is the reflection of \(y=\mathrm{f}(x)\) in the line \(y=x\). It has vertical asymptote \(x=3\) and \(x\)-intercept \((5,0)\).

For \(\mathrm{g}^{-1}\), write

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

Then

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

So

\(\mathrm{e}^x+2=\frac{3}{2-y},\)

and hence

\(x=\ln\left(\frac{3}{2-y}-2\right).\)

Interchanging \(x\) and \(y\),

\(\mathrm{g}^{-1}(x)=\ln\left(\frac{3}{2-x}-2\right).\)

The domain of \(\mathrm{g}^{-1}\) is the range of \(\mathrm{g}\). Since \(x\geqslant0\), \(\mathrm{g}(0)=1\), and as \(x\to\infty\), \(\mathrm{g}(x)\to2\) from below.

Therefore the domain is

\(1\leqslant x\lt 2.\)