(a) To find \(\frac{u}{v}\), multiply numerator and denominator by the conjugate of the denominator: \(\frac{-4 + 2i}{3 + i} \times \frac{3 - i}{3 - i} = \frac{(-4 + 2i)(3 - i)}{(3 + i)(3 - i)}\).

The denominator becomes \(3^2 - i^2 = 10\).

The numerator becomes \(-12 + 4i + 6i - 2i^2 = -12 + 10i + 2 = -10 + 10i\).

Thus, \(\frac{u}{v} = \frac{-10 + 10i}{10} = -1 + i\).

(b) The modulus \(r\) is \(\sqrt{(-1)^2 + 1^2} = \sqrt{2}\).

The argument \(\theta\) is \(\arctan\left(\frac{1}{-1}\right) = \frac{3}{4}\pi\).

Thus, \(\frac{u}{v} = \sqrt{2}e^{\frac{3}{4}\pi i}\).

\((c) Since C represents 2u + v, we have BC = 2OA. Both OA and BC are parallel.\)

(d) The angle AOB is given by \(\arg u - \arg v = \arg \left(\frac{u}{v}\right) = \frac{3}{4}\pi\).