(a) To find the square roots of the complex number \(1 - 2\sqrt{6}i\), denote the square root as \(x + iy\). Then, \((x + iy)^2 = 1 - 2\sqrt{6}i\).

Equate real and imaginary parts: \(x^2 - y^2 = 1\) and \(2xy = -2\sqrt{6}\).

From \(2xy = -2\sqrt{6}\), we have \(xy = -\sqrt{6}\).

Substitute \(y = -\frac{\sqrt{6}}{x}\) into \(x^2 - y^2 = 1\):

\(x^2 - \left(-\frac{\sqrt{6}}{x}\right)^2 = 1\)

\(x^2 - \frac{6}{x^2} = 1\)

Multiply through by \(x^2\):

\(x^4 - 6 = x^2\)

\(x^4 - x^2 - 6 = 0\)

Let \(u = x^2\), then \(u^2 - u - 6 = 0\).

Factorize: \((u - 3)(u + 2) = 0\)

\(u = 3\) or \(u = -2\) (discard \(u = -2\) as \(x^2\) cannot be negative).

\(x^2 = 3\), so \(x = \pm \sqrt{3}\).

Substitute back to find \(y\):

\(y = -\frac{\sqrt{6}}{x} = -\frac{\sqrt{6}}{\pm \sqrt{3}} = \mp \sqrt{2}\).

Thus, the square roots are \(\pm (\sqrt{3} - i\sqrt{2})\).

(b) The inequality \(|z - 3i| \leq 2\) represents a circle centered at \(3i\) with radius 2 on the Argand diagram.

Shade the interior of this circle.

The greatest value of \(\arg z\) occurs at the topmost point of the circle, which is at \(3 + 2i\).

Calculate \(\arg(3 + 2i)\):

\(\arctan\left(\frac{2}{3}\right)\)

\(\arg z = 131.8^\circ\) or \(2.30\) radians.