(a) In limiting equilibrium, the frictional force is equal to the applied force. The normal reaction \(R\) is given by \(R = 0.2g\).

The frictional force is \(\mu R\), so \(1.2 = \mu \times 0.2g\).

Solving for \(\mu\), we get \(\mu = \frac{1.2}{0.2g} = 0.6\).

(b) Using Newton's Second Law, the net force is \(1.2 - 0.3 \times 0.2g = 0.2a\).

Solving for \(a\), we get \(a = 3 \, \text{m/s}^2\).

To find the distance travelled in the third second, use the equation \(s = ut + \frac{1}{2}at^2\).

For the first 3 seconds, \(s_3 = 0 + \frac{1}{2} \times 3 \times 3^2 = 13.5 \, \text{m}\).

For the first 2 seconds, \(s_2 = 0 + \frac{1}{2} \times 3 \times 2^2 = 6 \, \text{m}\).

Thus, the distance travelled in the third second is \(13.5 - 6 = 7.5 \, \text{m}\).