Answer: range \(f\in\mathbb R\); \(f^{-1}(x)=\dfrac{e^{x/5}-3}{2}\), domain \(x\in\mathbb R\).

For \(f(x)=5\ln(2x+3)\), the domain is \(x\gt -\frac32\). As \(x\) approaches \(-\frac32\) from the right, \(2x+3\) approaches \(0\), so \(f(x)\) decreases without bound. As \(x\) increases, \(f(x)\) increases without bound. Hence the range is

\(f\in\mathbb R.\)

To find the inverse, let

\(y=5\ln(2x+3).\)

Then

\(\frac{y}{5}=\ln(2x+3),\)

so

\(e^{y/5}=2x+3.\)

Thus

\(x=\frac{e^{y/5}-3}{2}.\)

Therefore

\(f^{-1}(x)=\frac{e^{x/5}-3}{2}.\)

The domain of \(f^{-1}\) is the range of \(f\), so the domain is \(x\in\mathbb R\).

For the sketch, \(y=f(x)\) has vertical asymptote \(x=-\frac32\), \(x\)-intercept \((-1,0)\), and \(y\)-intercept \((0,5\ln3)\).

The graph of \(y=f^{-1}(x)\) is the reflection in \(y=x\), so it has horizontal asymptote \(y=-\frac32\), \(x\)-intercept \((5\ln3,0)\), and \(y\)-intercept \((0,-1)\).