Answer: \(g^{-1}(x)=\frac{1}{x-3}\), domain \(3\lt x\leq4\), range \(x\geq1\).

(a) Let

\(y=3+\frac1x.\)

Then

\(y-3=\frac1x,\)

so

\(x=\frac{1}{y-3}.\)

Therefore

\(\boxed{g^{-1}(x)=\frac{1}{x-3}}.\)

For \(x\geq1\), the value of \(\frac1x\) satisfies \(0\lt \frac1x\leq1\). Hence

\(3\lt g(x)\leq4.\)

So the domain of \(g^{-1}\) is

\(\boxed{3\lt x\leq4},\)

and the range of \(g^{-1}\) is

\(\boxed{x\geq1}.\)

(b) The graph of \(y=h^{-1}(x)\) is the reflection of the graph of \(y=h(x)\) in the line \(y=x\).

The graph of \(y=h(x)\) begins to the right of \(x=\frac23\), is increasing, and is concave down. Since \(y=h(x)\) intersects \(y=x\) at two distinct points, the graph of \(y=h^{-1}(x)\) passes through those same two points after reflection.