(i) Integrate by parts: Let \(u = x\) and \(dv = e^{-\frac{1}{2}x} \, dx\). Then \(du = dx\) and \(v = -2e^{-\frac{1}{2}x}\).

\(\int x e^{-\frac{1}{2}x} \, dx = -2xe^{-\frac{1}{2}x} + 2 \int e^{-\frac{1}{2}x} \, dx\).

\(= -2xe^{-\frac{1}{2}x} - 4e^{-\frac{1}{2}x} + C\).

Evaluate from 0 to \(a\):

\(\left[-2ae^{-\frac{1}{2}a} - 4e^{-\frac{1}{2}a}\right] - \left[-2(0)e^{-\frac{1}{2}(0)} - 4e^{-\frac{1}{2}(0)}\right] = 2\).

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

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

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

\(a+2 = e^{\frac{1}{2}a}\).

\(a = 2 \ln(a+2)\).

(ii) Calculate \(2 \ln(3+2) = 2 \ln 5 \approx 3.21888\) and \(2 \ln(3.5+2) = 2 \ln 5.5 \approx 3.58352\).

Since \(3.21888 < 3.5\) and \(3.58352 > 3\), \(a\) lies between 3 and 3.5.

(iii) Use iteration: \(a_{n+1} = 2 \ln(a_n + 2)\).

Start with \(a_0 = 3\).

\(a_1 = 2 \ln(3 + 2) = 3.21888\).

\(a_2 = 2 \ln(3.21888 + 2) = 3.3434\).

\(a_3 = 2 \ln(3.3434 + 2) = 3.3584\).

\(a_4 = 2 \ln(3.3584 + 2) = 3.3608\).

\(a_5 = 2 \ln(3.3608 + 2) = 3.3612\).

\(a_6 = 2 \ln(3.3612 + 2) = 3.3613\).

Thus, \(a \approx 3.36\) to 2 decimal places.