(a) To find the stationary points, we first find the derivative \(\frac{dy}{dx}\).

\(\frac{dy}{dx} = \left( \frac{5}{3} (4x - 3)^{\frac{2}{3}} \right) \times 4 - \frac{20}{3}\).

Setting \(\frac{dy}{dx} = 0\) gives:

\(\frac{20}{3} (4x - 3)^{\frac{2}{3}} = \frac{20}{3}\).

This leads to \((4x - 3)^{\frac{2}{3}} = 1\), so \(4x - 3 = \pm 1\).

Solving gives \(x = \frac{1}{2}\) and \(x = 1\).

To determine the nature, we find the second derivative \(\frac{d^2y}{dx^2}\).

\(\frac{d^2y}{dx^2} = \frac{40}{9} (4x - 3)^{-\frac{1}{3}} \times 4\).

For \(x = \frac{1}{2}\), \(\frac{d^2y}{dx^2} < 0\), so it is a maximum.

For \(x = 1\), \(\frac{d^2y}{dx^2} > 0\), so it is a minimum.

(b) The function \(f\) is increasing where \(\frac{dy}{dx} > 0\).

This occurs when \(x < \frac{1}{2}\) or \(x > 1\).