(a) Let \(w = a + bi\), then \(w^* = a - bi\). Substitute into the equation:

\(3(a + bi) + 2i(a - bi) = 17 + 8i\).

Expand and separate real and imaginary parts:

Real: \(3a - 2b = 17\)

Imaginary: \(3b + 2a = 8\)

Solve these simultaneous equations to find \(a\) and \(b\):

From \(3a - 2b = 17\), express \(b\) in terms of \(a\):

\(b = \frac{3a - 17}{2}\)

Substitute into \(3b + 2a = 8\):

\(3\left(\frac{3a - 17}{2}\right) + 2a = 8\)

\(\frac{9a - 51}{2} + 2a = 8\)

\(9a - 51 + 4a = 16\)

\(13a = 67\)

\(a = \frac{67}{13}\)

Substitute \(a\) back to find \(b\):

\(b = \frac{3\left(\frac{67}{13}\right) - 17}{2}\)

\(b = -2\)

Thus, \(w = 7 - 2i\).

(b) The locus \(|z - 3| = |z - 3i|\) is the perpendicular bisector of the line segment joining \(3\) and \(3i\), which is the line \(y = x\).

The locus \(\arg(z - 2i) = \frac{1}{6}\pi\) is a line making an angle \(\frac{1}{6}\pi\) with the positive real axis, passing through \(2i\).

Find the intersection of \(y = x\) and \(\arg(z - 2i) = \frac{1}{6}\pi\):

\(y = x\) implies \(z = x + xi\).

\(\arg((x + xi) - 2i) = \frac{1}{6}\pi\)

\(\arg(x + (x - 2)i) = \frac{1}{6}\pi\)

\(\arctan\left(\frac{x - 2}{x}\right) = \frac{1}{6}\pi\)

Solve for \(x\):

\(\frac{x - 2}{x} = \tan\left(\frac{1}{6}\pi\right)\)

\(\frac{x - 2}{x} = \frac{1}{\sqrt{3}}\)

\(x - 2 = \frac{x}{\sqrt{3}}\)

\(x(1 - \frac{1}{\sqrt{3}}) = 2\)

\(x = \frac{2\sqrt{3}}{\sqrt{3} - 1}\)

\(x = 6.69\) (to 3 significant figures)

Thus, \(z = 6.69 + 6.69i\).

Express \(z\) in polar form:

\(r = \sqrt{6.69^2 + 6.69^2} = 6.69\)

\(\theta = \frac{\pi}{4}\)

So, \(z = 6.69e^{\frac{\pi}{4}i}\).