Answer: Range of \(f\): \(y\gt 1\); range of \(g\): \(\mathbb{R}\); \(fg^2(0)=3e^4+1\approx165\). Intercepts: \((0,4)\) and \((4,0)\).

Since \(e^{2x}\gt 0\) for all real \(x\),

\(3e^{2x}+1\gt 1.\)

So the range of \(f\) is

\(y\gt 1.\)

The function \(g(x)=x+1\) is linear with domain \(\mathbb{R}\), so its range is

\(\mathbb{R}.\)

Next, \(g^2(0)\) means \(g(g(0))\). Since

\(g(0)=1,\qquad g(1)=2,\)

we have

\(fg^2(0)=f(2)=3e^4+1\approx165.\)

For the sketch, \(y=f(x)\) has horizontal asymptote \(y=1\) and meets the y-axis at

\(f(0)=4,\)

so the point is \((0,4)\).

The graph of \(y=f^{-1}(x)\) is the reflection of \(y=f(x)\) in the line \(y=x\). Therefore it meets the x-axis at the reflected point

\((4,0).\)