(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 \(v = -\frac{1}{2}e^{-2x}\).

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

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

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

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

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

To solve \(\int_{\frac{1}{3}\pi}^{\pi} x \sin \frac{1}{2}x \, dx\), we use integration by parts.

Let \(u = x\) and \(dv = \sin \frac{1}{2}x \, dx\). Then \(du = dx\) and \(v = -2 \cos \frac{1}{2}x\).

Applying integration by parts:

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

\(= -2x \cos \frac{1}{2}x + 2 \int \cos \frac{1}{2}x \, dx\)

Integrate \(\cos \frac{1}{2}x\):

\(\int \cos \frac{1}{2}x \, dx = 4 \sin \frac{1}{2}x\)

Thus, the integral becomes:

\(-2x \cos \frac{1}{2}x + 4 \sin \frac{1}{2}x\)

Evaluate from \(\frac{1}{3}\pi\) to \(\pi\):

\(\left[ -2x \cos \frac{1}{2}x + 4 \sin \frac{1}{2}x \right]_{\frac{1}{3}\pi}^{\pi}\)

Calculate at \(x = \pi\):

\(-2\pi \cos \frac{1}{2}\pi + 4 \sin \frac{1}{2}\pi = 4\)

Calculate at \(x = \frac{1}{3}\pi\):

\(-2 \left( \frac{1}{3}\pi \right) \cos \frac{1}{6}\pi + 4 \sin \frac{1}{6}\pi\)

\(= -\frac{2}{3}\pi \cdot \frac{\sqrt{3}}{2} + 4 \cdot \frac{1}{2}\)

\(= -\frac{\sqrt{3}}{3}\pi + 2\)

Subtract the two results:

\(4 - \left( -\frac{\sqrt{3}}{3}\pi + 2 \right)\)

\(= 2 + \frac{\sqrt{3}}{3}\pi\)

To solve \(\int_{0}^{1} (1-x)e^{-\frac{1}{2}x} \, dx\), we use integration by parts. Let \(u = 1-x\) and \(dv = e^{-\frac{1}{2}x} \, dx\).

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

Applying integration by parts:

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

\(= (1-x)(-2e^{-\frac{1}{2}x}) \bigg|_{0}^{1} + 2 \int_{0}^{1} e^{-\frac{1}{2}x} \, dx\)

Evaluating the first term:

\(= [-2(1-x)e^{-\frac{1}{2}x}] \bigg|_{0}^{1}\)

\(= [-2(0)e^{-\frac{1}{2}(1)} + 2(1)e^{0}]\)

\(= [0 + 2] = 2\)

Now, evaluate the second integral:

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

\(= 2 \left[ -2e^{-\frac{1}{2}} + 2 \right]\)

\(= -4e^{-\frac{1}{2}} + 4\)

Combining both parts:

\(2 - 4e^{-\frac{1}{2}} + 4 = 4 - 4e^{-\frac{1}{2}}\)

Thus, the integral evaluates to:

\(4e^{-\frac{1}{2}} - 2\)

To solve \(\int_{0}^{\pi} x^2 \sin x \, dx\), we use integration by parts. Let \(u = x^2\) and \(dv = \sin x \, dx\). Then \(du = 2x \, dx\) and \(v = -\cos x\).

Applying integration by parts:

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

\(\int x^2 \sin x \, dx = -x^2 \cos x + \int 2x \cos x \, dx\)

Now, integrate \(\int 2x \cos x \, dx\) by parts again. Let \(u = 2x\) and \(dv = \cos x \, dx\). Then \(du = 2 \, dx\) and \(v = \sin x\).

\(\int 2x \cos x \, dx = 2x \sin x - \int 2 \sin x \, dx\)

\(= 2x \sin x + 2 \cos x\)

Substituting back, we have:

\(\int x^2 \sin x \, dx = -x^2 \cos x + 2x \sin x + 2 \cos x\)

Evaluate from 0 to \(\pi\):

\(\left[ -x^2 \cos x + 2x \sin x + 2 \cos x \right]_{0}^{\pi}\)

At \(x = \pi\):

\(-\pi^2 \cos \pi + 2\pi \sin \pi + 2 \cos \pi = \pi^2 + 2(-1) = \pi^2 - 2\)

At \(x = 0\):

\(-0^2 \cos 0 + 2 \cdot 0 \cdot \sin 0 + 2 \cos 0 = 2\)

Subtracting, we get:

\((\pi^2 - 2) - 2 = \pi^2 - 4\)

Let \(u = \ln x\) and \(dv = dx\). Then \(du = \frac{1}{x} \, dx\) and \(v = x\).

Using integration by parts, \(\int u \, dv = uv - \int v \, du\).

Substitute the values: \(\int \ln x \, dx = x \ln x - \int x \cdot \frac{1}{x} \, dx\).

This simplifies to \(x \ln x - \int 1 \, dx = x \ln x - x + C\).

Evaluate from 2 to 4:

\(\left[ x \ln x - x \right]_{2}^{4} = (4 \ln 4 - 4) - (2 \ln 2 - 2)\).

Calculate each term:

\(4 \ln 4 - 4 = 4 \cdot 2 \ln 2 - 4 = 8 \ln 2 - 4\).

\(2 \ln 2 - 2 = 2 \ln 2 - 2\).

Combine the results:

\((8 \ln 2 - 4) - (2 \ln 2 - 2) = 8 \ln 2 - 4 - 2 \ln 2 + 2\).

\(= 6 \ln 2 - 2\).