(i) To solve \(\int_0^a x \cos \frac{1}{3}x \, dx = 3\), use integration by parts:

Let \(u = x\) and \(dv = \cos \frac{1}{3}x \, dx\). Then \(du = dx\) and \(v = 3 \sin \frac{1}{3}x\).

Integration by parts gives:

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

\(= x(3 \sin \frac{1}{3}x) - \int 3 \sin \frac{1}{3}x \, dx\)

\(= 3x \sin \frac{1}{3}x + 9 \cos \frac{1}{3}x\).

Substitute limits and equate to 3:

\(3a \sin \frac{a}{3} + 9 \cos \frac{a}{3} - 9 = 3\)

\(3a \sin \frac{a}{3} + 9 \cos \frac{a}{3} = 12\)

\(a \sin \frac{a}{3} = 4 - 3 \cos \frac{a}{3}\)

\(a = \frac{4 - 3 \cos \frac{a}{3}}{\sin \frac{a}{3}}\)

(ii) Calculate values at \(a = 2.5\) and \(a = 3\):

\(f(a) = a \sin \frac{a}{3} + 3 \cos \frac{a}{3} - 4\)

\(f(2.5) = 0.179\)

\(f(3) = -0.173\)

Since \(f(2.5) > 0\) and \(f(3) < 0\), \(a\) lies between 2.5 and 3.

(iii) Use the iterative formula:

\(a_{n+1} = \frac{4 - 3 \cos \frac{1}{3}a_n}{\sin \frac{1}{3}a_n}\)

Starting with \(a_0 = 2.75\):

\(a_1 = 2.73554\)

\(a_2 = 2.73599\)

\(a_3 = 2.73600\)

\(a_4 = 2.73600\)

Thus, \(a \approx 2.736\) correct to 3 decimal places.