(i) Multiply the numerator and denominator of \(\frac{3+i}{2-i}\) by the conjugate of the denominator, \(2+i\):

\(\frac{(3+i)(2+i)}{(2-i)(2+i)} = \frac{6 + 3i + 2i + i^2}{4 + 1} = \frac{5 + 5i}{5} = 1 + i\)

Thus, \(u = 1 + i\).

(ii) The modulus of \(u = 1 + i\) is \(\sqrt{1^2 + 1^2} = \sqrt{2}\).

The argument of \(u = 1 + i\) is \(\arctan\left(\frac{1}{1}\right) = 45^{\circ}\) or \(\frac{\pi}{4}\).

(iii) On the Argand diagram, plot the point \(u = 1 + i\). Draw a circle centered at \(u\) with radius 1, representing the locus \(|z-u| = 1\).

(iv) The least value of \(|z|\) for points on this locus is the distance from the origin to the closest point on the circle. The center of the circle is at \(1 + i\) with radius 1, so the least distance is \(\sqrt{2} - 1\).