Answer: The length of the curve is \(\mathrm{e}^2+\frac{5}{3}\).

For a parametric curve, the length from \(t=0\) to \(t=2\) is

\(\displaystyle s=\int_0^2 \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\,dt.\)

Here

\(x=\mathrm{e}^t-\frac13 t^3\), so

\(\displaystyle \frac{dx}{dt}=\mathrm{e}^t-t^2.\)

Also

\(y=4\mathrm{e}^{t/2}(t-2)\), so using the product rule,

\(\displaystyle \frac{dy}{dt}=4\left(\frac12\mathrm{e}^{t/2}(t-2)+\mathrm{e}^{t/2}\right)=2t\mathrm{e}^{t/2}.\)

Hence

\(\displaystyle \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}=\sqrt{\left(\mathrm{e}^t-t^2\right)^2+\left(2t\mathrm{e}^{t/2}\right)^2}.\)

Simplifying,

\(\displaystyle \left(\mathrm{e}^t-t^2\right)^2+4t^2\mathrm{e}^t=\mathrm{e}^{2t}-2t^2\mathrm{e}^t+t^4+4t^2\mathrm{e}^t=\mathrm{e}^{2t}+2t^2\mathrm{e}^t+t^4=(\mathrm{e}^t+t^2)^2.\)

So

\(\displaystyle \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}=\mathrm{e}^t+t^2.\)

Therefore

\(\displaystyle s=\int_0^2 (\mathrm{e}^t+t^2)\,dt=\left[\mathrm{e}^t+\frac13 t^3\right]_0^2=\left(\mathrm{e}^2+\frac83\right)-1=\mathrm{e}^2+\frac53.\)

So the length of the curve is \(\boxed{\mathrm{e}^2+\frac53}\).