Answer: \(\left(\frac{11}{20},-\frac{11e^2}{6}\right)\).

Rewrite the exponential terms in a common form, or introduce a substitution, so that the equation becomes algebraic.

\(y=e^x(2x+5)^{1/2}\).

Differentiate using the product rule:

\(\frac{\mathrm dy}{\mathrm dx}=e^x(2x+5)^{1/2}+e^x(2x+5)^{-1/2}\).

At \(x=2\), \(y=3e^2\), and

\(\frac{\mathrm dy}{\mathrm dx}=e^2\left(3+\frac13\right)=\frac{10e^2}{3}\).

The tangent is therefore

\(y-3e^2=\frac{10e^2}{3}(x-2)\).

For the \(x\)-intercept, put \(y=0\):

\(-3e^2=\frac{10e^2}{3}(x-2)\), giving \(x=\frac{11}{10}\). So \(X=\left(\frac{11}{10},0\right)\).

For the \(y\)-intercept, put \(x=0\):

\(y-3e^2=-\frac{20e^2}{3}\), so \(y=-\frac{11e^2}{3}\). Hence \(Y=\left(0,-\frac{11e^2}{3}\right)\).

The midpoint of \(XY\) is

\(\left(\frac{11}{20},-\frac{11e^2}{6}\right)\).

This gives the required answer: \(\left(\frac{11}{20},-\frac{11e^2}{6}\right)\).