Answer: (a)(i) \(f(x)\leq -1\); (a)(ii) \(x=-2\); (a)(iii) \(\frac{\pi}{2}\lt x\lt \frac{3\pi}{2}\); (a)(iv) \(x=\frac{5\pi}{6},\frac{7\pi}{6}\).

For an inverse function, interchange \(x\) and \(y\), then rearrange to make \(y\) the subject. The domain of the original function becomes the range of the inverse.

(a)(i) For

\(\frac{\pi}{2}\lt x\lt \frac{3\pi}{2},\)

we have \(\cos x\lt 0\). Therefore

\(\operatorname{sec} x=\frac{1}{\cos x}\leq -1.\)

So the range of \(f\) is

\(f(x)\leq -1.\)

(a)(ii) If

\(f^{-1}(x)=\frac{2\pi}{3},\)

then

\(x=f\left(\frac{2\pi}{3}\right).\)

Hence

\(x=\operatorname{sec}\frac{2\pi}{3}.\)

Since

\(\cos\frac{2\pi}{3}=-\frac12,\)

we get

\(x=-2.\)

(a)(iii) Since \(g\) is defined for all real inputs, the domain of \(gf\) is just the domain of \(f\):

\(\frac{\pi}{2}\lt x\lt \frac{3\pi}{2}.\)

(a)(iv) Now

\(gf(x)=g(f(x))=3\bigl((\operatorname{sec} x)^2-1\bigr).\)

Using \(\operatorname{sec}^2x-1=\tan^2x\),

\(gf(x)=3\tan^2x.\)

So \(gf(x)=1\) gives

\(3\tan^2x=1.\)

Therefore

\(\tan^2x=\frac13.\)

In the interval \(\frac{\pi}{2}\lt x\lt \frac{3\pi}{2}\), the solutions are

\(x=\frac{5\pi}{6} \quad\text{and}\quad x=\frac{7\pi}{6}.\)

(b) For

\(h(x)=\ln(4-x),\)

the vertical asymptote is

\(x=4.\)

The intercepts of \(y=h(x)\) are:

\(y\text{-intercept: }(0,\ln4),\)

and

\(x\text{-intercept: }(3,0).\)

To find the inverse, let

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

Then

\(e^y=4-x,\)

so

\(x=4-e^y.\)

Thus

\(h^{-1}(x)=4-e^x.\)

The graph of \(y=h^{-1}(x)\) has horizontal asymptote

\(y=4.\)

Its intercepts are

\((0,3) \quad\text{and}\quad (\ln4,0).\)

The two graphs are reflections of each other in the line \(y=x\).