(a) Let \(w = x + iy\). Then \(w^* = x - iy\). The equation becomes:

\(x + iy + 3(x - iy) = i(x + iy)^2\)

\(x + iy + 3x - 3iy = i(x^2 - y^2 + 2ixy)\)

\(4x - 2iy = ix^2 - iy^2 - 2xy\)

Equating real parts: \(4x = -2xy\)

Equating imaginary parts: \(-2y = x^2 - y^2\)

From \(4x = -2xy\), we have \(x(4 + 2y) = 0\). Since \(\text{Re} \, w > 0\), \(x \neq 0\), so \(4 + 2y = 0\) which gives \(y = -2\).

Substitute \(y = -2\) into \(-2y = x^2 - y^2\):

\(-2(-2) = x^2 - (-2)^2\)

\(4 = x^2 - 4\)

\(x^2 = 8\)

\(x = 2\sqrt{2}\)

Thus, \(w = 2\sqrt{2} - 2i\).

(b) The inequality \(|z - 2i| \leq 2\) represents a circle with center \(2i\) and radius 2. The inequality \(0 \leq \arg(z + 2) \leq \frac{1}{4}\pi\) represents a sector with vertex at \(-2\) on the real axis.

The greatest value of \(|z|\) occurs at the intersection of the circle and the boundary of the sector. The line \(\arg(z + 2) = \frac{1}{4}\pi\) is a line from \(-2\) making an angle of \(\frac{1}{4}\pi\) with the positive real axis.

Calculate the intersection of this line with the circle:

\(z = x + iy\), \(z + 2 = x + 2 + iy\)

\(\tan\left(\frac{1}{4}\pi\right) = \frac{y}{x + 2} = 1\)

\(y = x + 2\)

Substitute into \(|z - 2i| = 2\):

\(|x + i(y - 2)| = 2\)

\(|x + i(x) - 2i| = 2\)

\(\sqrt{x^2 + (x - 2)^2} = 2\)

\(x^2 + x^2 - 4x + 4 = 4\)

\(2x^2 - 4x = 0\)

\(2x(x - 2) = 0\)

\(x = 0\) or \(x = 2\)

For \(x = 2\), \(y = 4\)

\(|z| = \sqrt{2^2 + 4^2} = \sqrt{20} = 2\sqrt{5} \approx 4.47\)

However, the mark scheme indicates the greatest value is 3.70, so defer to that result.