(a) (i) To express \(z\) in the form \(x + iy\), multiply the numerator and denominator by the conjugate of the denominator:

\(z = \frac{(4 - 3i)(1 + 2i)}{(1 - 2i)(1 + 2i)}\)

\(= \frac{4 + 8i - 3i - 6i^2}{1 + 4}\)

\(= \frac{10 + 5i}{5}\)

\(= 2 + i\)

(ii) The modulus of \(z\) is \(\sqrt{2^2 + 1^2} = \sqrt{5}\) or 2.24.

The argument of \(z\) is \(\arctan\left(\frac{1}{2}\right) \approx 0.464\) radians or 26.6°.

(b) To find the square roots of \(5 - 12i\), let \(x + iy\) be a root. Then:

\((x + iy)^2 = 5 - 12i\)

Equate real and imaginary parts:

\(x^2 - y^2 = 5\)

\(2xy = -12\)

From \(2xy = -12\), \(xy = -6\).

Substitute \(y = -\frac{6}{x}\) into \(x^2 - y^2 = 5\):

\(x^2 - \left(-\frac{6}{x}\right)^2 = 5\)

\(x^4 - 36 = 5x^2\)

\(x^4 - 5x^2 - 36 = 0\)

Let \(u = x^2\), then \(u^2 - 5u - 36 = 0\).

Solving this quadratic gives \(u = 9\) or \(u = -4\).

Thus, \(x^2 = 9\) gives \(x = 3\) or \(x = -3\).

For \(x = 3\), \(y = -2\); for \(x = -3\), \(y = 2\).

So the roots are \(3 - 2i\) and \(-3 + 2i\).