(a) Let the square root of \(7 - (6\sqrt{2})i\) be \(x + iy\). Then, \((x + iy)^2 = 7 - 6\sqrt{2}i\).

Expanding, we have \(x^2 - y^2 + 2xyi = 7 - 6\sqrt{2}i\).

Equating real and imaginary parts, we get:

\(x^2 - y^2 = 7\)

\(2xy = -6\sqrt{2}\)

From \(2xy = -6\sqrt{2}\), we have \(xy = -3\sqrt{2}\).

Substitute \(y = \frac{-3\sqrt{2}}{x}\) into \(x^2 - y^2 = 7\):

\(x^2 - \left(\frac{-3\sqrt{2}}{x}\right)^2 = 7\)

\(x^2 - \frac{18}{x^2} = 7\)

Multiply through by \(x^2\):

\(x^4 - 7x^2 - 18 = 0\)

Solving this quadratic in \(x^2\), we find \(x^2 = 9\) or \(x^2 = -2\).

Thus, \(x = \pm 3\) and \(y = \pm i\sqrt{2}\).

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

(b)(i) The locus \(|w - 1 - 2i| = 1\) is a circle with center \((1, 2)\) and radius 1.

The locus \(\text{arg}(z - 1) = \frac{3}{4}\pi\) is a half-line starting from \((1, 0)\) making an angle of \(\frac{3}{4}\pi\) with the positive real axis.

(b)(ii) The least value of \(|w - z|\) is the perpendicular distance from the center of the circle \((1, 2)\) to the line \(\text{arg}(z - 1) = \frac{3}{4}\pi\).

This distance is \(\sqrt{2} - 1\) (or 0.414).