Answer: (a) \(3\) s. (b) \(3\) m.

Since the acceleration is \(-6\text{ m s}^{-2}\) and the initial velocity is \(18\text{ m s}^{-1}\),

\(v=18-6t.\)

(a) The particle is instantaneously at rest when \(v=0\):

\(18-6t=0.\)

Hence

\(\boxed{t=3\text{ s}}.\)

(b) The third second is the interval from \(t=2\) to \(t=3\). Since the velocity is positive on this interval, the distance travelled is

\(\int_2^3(18-6t)\,dt.\)

So

\(\int_2^3(18-6t)\,dt=\left[18t-3t^2\right]_2^3.\)

Evaluate:

\((54-27)-(36-12)=27-24=3.\)

Therefore the distance travelled in the third second is

\(\boxed{3\text{ m}}.\)