To solve the problem, we follow these steps:

Part (i): Find \(\sin(a - 30^\circ)\)

We know \(\cos a = \frac{3}{5}\). Using the Pythagorean identity, \(\sin a = \sqrt{1 - \cos^2 a} = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \frac{4}{5}\).

Using the angle subtraction formula for sine:

\(\sin(a - 30^\circ) = \sin a \cos 30^\circ - \cos a \sin 30^\circ\)

Substitute the known values:

\(\sin(a - 30^\circ) = \frac{4}{5} \cdot \frac{\sqrt{3}}{2} - \frac{3}{5} \cdot \frac{1}{2}\)

\(= \frac{4\sqrt{3}}{10} - \frac{3}{10}\)

\(= \frac{1}{10}(4\sqrt{3} - 3)\)

Part (ii): Find \(\tan 2a\) and \(\tan 3a\)

Using the double angle formula for tangent:

\(\tan 2a = \frac{2 \tan a}{1 - \tan^2 a}\)

First, find \(\tan a = \frac{\sin a}{\cos a} = \frac{4/5}{3/5} = \frac{4}{3}\).

Substitute into the formula:

\(\tan 2a = \frac{2 \cdot \frac{4}{3}}{1 - \left(\frac{4}{3}\right)^2} = \frac{\frac{8}{3}}{1 - \frac{16}{9}}\)

\(= \frac{\frac{8}{3}}{-\frac{7}{9}} = -\frac{24}{7}\)

Using the angle addition formula for tangent:

\(\tan 3a = \tan(2a + a) = \frac{\tan 2a + \tan a}{1 - \tan 2a \tan a}\)

Substitute \(\tan 2a = -\frac{24}{7}\) and \(\tan a = \frac{4}{3}\):

\(\tan 3a = \frac{-\frac{24}{7} + \frac{4}{3}}{1 - \left(-\frac{24}{7}\right) \left(\frac{4}{3}\right)}\)

\(= \frac{-\frac{72}{21} + \frac{28}{21}}{1 + \frac{96}{21}}\)

\(= \frac{-\frac{44}{21}}{\frac{117}{21}} = -\frac{44}{117}\)