(i) Differentiate \(f(x) = 3x e^{-2x}\) using the product rule:

\(f'(x) = 3e^{-2x} - 6xe^{-2x}\).

Substitute \(x = -\frac{1}{2}\):

\(f'\left(-\frac{1}{2}\right) = 3e - 6\left(-\frac{1}{2}\right)e = 6e\).

(ii) Use integration by parts for \(\int_{-\frac{1}{2}}^{0} 3x e^{-2x} \, dx\):

Let \(u = 3x\) and \(dv = e^{-2x} \, dx\).

Then \(du = 3 \, dx\) and \(v = -\frac{1}{2}e^{-2x}\).

\(\int u \, dv = uv - \int v \, du\).

\(= \left[-\frac{3}{2}xe^{-2x}\right]_{-\frac{1}{2}}^{0} + \frac{3}{2}\int_{-\frac{1}{2}}^{0} e^{-2x} \, dx\).

\(= \left[-\frac{3}{2}(0)e^{0} + \frac{3}{2}\left(-\frac{1}{2}e\right)e^{1}\right] + \frac{3}{2}\left[-\frac{1}{2}e^{-2x}\right]_{-\frac{1}{2}}^{0}\).

\(= 0 + \frac{3}{4}e - \frac{3}{4}\).

\(= -\frac{3}{4}\).