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

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

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

Substitute limits and equate to 6: \(2ae^{\frac{1}{2}a} - 4e^{\frac{1}{2}a} = 6\)

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

(ii) Sketch the graphs of \(y = x\) and \(y = 2 + e^{-\frac{1}{2}x}\). They intersect at one point, indicating one root.

(iii) Check the sign of \(x - e^{-\frac{1}{2}x} - 2\) at \(x = 2\) and \(x = 2.5\). The sign changes, confirming a root between these values.

(iv) Use the iterative formula \(x_{n+1} = 2 + e^{-\frac{1}{2}x_n}\) with initial guess \(x_0 = 2\).

Calculate iterations: \(x_1 = 2.3033\), \(x_2 = 2.3113\), \(x_3 = 2.3105\), \(x_4 = 2.3107\).

The value of a is 2.31 to 2 decimal places.