(a) To solve the equations \(u + 2v = 2i\) and \(iu + v = 3\), we can express \(u\) and \(v\) in terms of real and imaginary parts. Let \(u = x + yi\) and \(v = a + bi\).

Substitute into the first equation:

\((x + yi) + 2(a + bi) = 2i\)

Equating real and imaginary parts gives:

\(x + 2a = 0\)

\(y + 2b = 2\)

Substitute into the second equation:

\(i(x + yi) + (a + bi) = 3\)

\((-y + xi) + (a + bi) = 3\)

Equating real and imaginary parts gives:

\(a - y = 3\)

\(x + b = 0\)

Solving these equations, we find:

\(x = -2, y = -2, a = 1, b = 2\)

Thus, \(u = -2 - 2i\) and \(v = 1 + 2i\).

(b) The locus \(|z + i| = 1\) is a circle centered at \(-i\) with radius 1. The locus \(\text{arg}(w - 2) = \frac{3}{4}\pi\) is a half-line starting from 2 at an angle of \(\frac{3}{4}\pi\) to the real axis.

The least value of \(|z - w|\) is found by considering the distance from the center of the circle to the line, which is \(\frac{3}{\sqrt{2} - 1}\) or approximately 1.12.

(i) For \(|z|^2 = zz*\), we have \(z = a + ib\) and \(z* = a - ib\). Thus, \(|z|^2 = (a + ib)(a - ib) = a^2 + b^2\).

For \((z - ki)* = z* + ki\), we have \(z - ki = a + ib - ki\). The conjugate is \((a + ib - ki)^* = a - ib + ki\).

(ii) Start with \(|z - 10i| = 2|z - 4i|\). Square both sides: \((z - 10i)(z^* + 10i) = 4(z - 4i)(z^* + 4i)\).

Expand and simplify to get \(zz* - 2iz* + 2iz - 12 = 0\).

Rearrange to find \(|z - 2i| = 4\).

(iii) The equation \(|z - 2i| = 4\) represents a circle in the Argand diagram with center at \(2i\) and radius 4.

(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.

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

The inequality \(|z - 3 - i| \leq 3\) represents a circle centered at \((3, 1)\) with radius 3 on the Argand diagram.

The inequality \(|z| \geq |z - 4i|\) represents the region above the line \(y = 2\) (since \(|z|\) is the distance from the origin and \(|z - 4i|\) is the distance from the point \((0, 4)\), the line \(y = 2\) is the perpendicular bisector of the segment joining these points).

Therefore, the solution is the region inside the circle centered at \((3, 1)\) with radius 3, and above the line \(y = 2\).