Answer: range of \(f\): \(f\gt3\); range of \(g\): \(\mathbb R\); \(x=e^9+3\); \(x=2\).

(a) Since \(e^x\gt0\) for all real \(x\),

\(f(x)=3+e^x\gt3.\)

So the range of \(f\) is

\(f\gt3.\)

The function \(g(x)=9x-5\) is linear with non-zero gradient and domain \(\mathbb R\). Therefore its range is

\(\mathbb R.\)

(b) To find \(f^{-1}\), let

\(y=3+e^x.\)

Then

\(e^x=y-3,\)

so

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

Hence

\(f^{-1}(x)=\ln(x-3).\)

Also

\(g'(x)=9.\)

So

\(\ln(x-3)=9.\)

Therefore

\(x-3=e^9,\)

and hence

\(x=e^9+3.\)

(c) Here \(g^2(x)\) means \(g(g(x))\). Since \(g(x)=9x-5\),

\(g(g(x))=9(9x-5)-5.\)

So

\(9(9x-5)-5=112.\)

Thus

\(81x-50=112.\)

Hence

\(81x=162,\)

and so

\(x=2.\)