(i) To find the expression for \(s\), we start by integrating the acceleration to find the velocity:

\(v = \int (6t - 12) \, dt = 3t^2 - 12t + C\)

Next, integrate the velocity to find the displacement:

\(s = \int (3t^2 - 12t + C) \, dt = t^3 - 6t^2 + Ct + D\)

Using the conditions \(s = 5\) when \(t = 1\) and \(s = 1\) when \(t = 3\), we set up the equations:

\(5 = 1^3 - 6(1)^2 + C(1) + D\)

\(1 = 3^3 - 6(3)^2 + 3C + D\)

Solving these gives \(C = 9\) and \(D = 1\). Thus, \(s = t^3 - 6t^2 + 9t + 1\).

(ii) The particle is at instantaneous rest when \(v = 0\):

\(3t^2 - 12t + 9 = 0\)

\(3(t^2 - 4t + 3) = 0\)

\((t - 1)(t - 3) = 0\)

Thus, \(t = 1\) or \(t = 3\).

(iii) To find the total distance travelled, evaluate \(s\) at key points:

For \(0 < t < 1\), \(s = (1 - 6 + 9 + 1) - 1 = 4\)

For \(1 < t < 3\), \(s = (27 - 54 + 27 + 1) - 5 = 4\)

For \(3 < t \leq 4\), \(s = (64 - 96 + 36 + 1) - 1 = 4\)

Total distance is \(4 + 4 + 4 = 12\) m.