(a) In limiting equilibrium, the frictional force \(F\) is equal to \(\mu R\), where \(R\) is the normal reaction force. The forces acting on the particle are its weight \(0.4g\) and the normal reaction \(R\). The component of the weight perpendicular to the plane is \(0.4g \cos 30^{\circ}\), and the component parallel to the plane is \(0.4g \sin 30^{\circ}\).

Since the particle is in limiting equilibrium, \(F = \mu R = 0.4g \sin 30^{\circ}\).

The normal reaction \(R = 0.4g \cos 30^{\circ}\).

Thus, \(\mu = \frac{0.4g \sin 30^{\circ}}{0.4g \cos 30^{\circ}} = \frac{\sin 30^{\circ}}{\cos 30^{\circ}} = \tan 30^{\circ} = \frac{1}{3} \sqrt{3}\).

(b) Applying Newton's second law along the plane: \(7.2 - 0.4g \sin 30^{\circ} - F = 0.4a\).

Substituting \(F = \mu R = \frac{1}{3} \sqrt{3} \times 0.4g \cos 30^{\circ}\), we find \(a = 8 \text{ m/s}^2\).

Using the equation of motion \(s = ut + \frac{1}{2}at^2\), where \(s = 1 \text{ m}\), \(u = 0\), and \(a = 8 \text{ m/s}^2\), we solve for \(t\):

\(1 = 0 + \frac{1}{2} \times 8 \times t^2\)

\(t^2 = \frac{1}{4}\)

\(t = 0.5 \text{ s}\).