(i) To find the displacement, integrate the velocity function \(v = 20t - t^3\) with respect to \(t\):

\(s(t) = \int (20t - t^3) \, dt = 10t^2 - 0.25t^4 + C\)

Given that at \(t = 0\), \(s = -36\), we find \(C\):

\(-36 = 10(0)^2 - 0.25(0)^4 + C\)

\(C = -36\)

Thus, the displacement is \(s(t) = 10t^2 - 0.25t^4 - 36\).

(ii) Substitute \(t = 4\) into the displacement expression:

\(s(4) = 10(4)^2 - 0.25(4)^4 - 36\)

\(s(4) = 160 - 64 - 36 = 60\)

Therefore, the displacement is 60 m.

(iii) To find when the particle is at \(O\), set \(s(t) = 0\):

\(10t^2 - 0.25t^4 - 36 = 0\)

Factor the equation:

\((t^2 - 36)(1 - 0.25t^2) = 0\)

Solving \(t^2 - 36 = 0\) gives \(t = 6\) or \(t = -6\).

Solving \(1 - 0.25t^2 = 0\) gives \(t = 2\) or \(t = -2\).

Since \(t\) must be positive, the values are \(t = 2\) and \(t = 6\).