Answer: (a) \(x=\ln\left(\frac3{\sqrt2}-1\right)\); (b) \(\mathrm f^{-1}\) does not exist; \(\mathrm g^{-1}(x)=\ln(x-1)\), domain \(x\gt1\).

We start with the main method. Check whether the function is one-to-one on the stated domain before finding an inverse.

(a) Since \(\mathrm{fg}(x)=\mathrm f(\mathrm g(x))\),

\(\mathrm{fg}(x)=2(\mathrm e^x+1)^2-1.\)

Set this equal to \(8\):

\(2(\mathrm e^x+1)^2-1=8.\)

Then

\(2(\mathrm e^x+1)^2=9,\)

so

\((\mathrm e^x+1)^2=\frac92.\)

Because \(\mathrm e^x+1\gt0\),

\(\mathrm e^x+1=\frac3{\sqrt2}.\)

Therefore

\(\mathrm e^x=\frac3{\sqrt2}-1.\)

So

\(x=\ln\left(\frac3{\sqrt2}-1\right).\)

(b) The function \(\mathrm f(x)=2x^2-1\) is not one-to-one for all real values of \(x\). For example, \(\mathrm f(1)=\mathrm f(-1)\).

Therefore \(\mathrm f^{-1}\) does not exist on the stated domain.

For \(\mathrm g\), let

\(y=\mathrm e^x+1.\)

Then

\(y-1=\mathrm e^x.\)

Taking natural logarithms gives

\(x=\ln(y-1).\)

So

\(\mathrm g^{-1}(x)=\ln(x-1).\)

Since the input to the logarithm must be positive, the domain is

\(x\gt1.\)

This completes the solution and gives the required result.