(i) To find the velocity \(v\), differentiate the displacement \(s\) with respect to \(t\):

\(v = \frac{ds}{dt} = \frac{d}{dt}(t^3 - 4t^2 + 4t) = 3t^2 - 8t + 4\).

(ii) To find when the particle is at instantaneous rest, set \(v = 0\):

\(3t^2 - 8t + 4 = 0\).

Solving this quadratic equation using the quadratic formula \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 3\), \(b = -8\), \(c = 4\):

\(t = \frac{8 \pm \sqrt{(-8)^2 - 4 \times 3 \times 4}}{2 \times 3} = \frac{8 \pm \sqrt{64 - 48}}{6} = \frac{8 \pm \sqrt{16}}{6} = \frac{8 \pm 4}{6}\).

This gives \(t = \frac{12}{6} = 2\) and \(t = \frac{4}{6} = \frac{2}{3}\).

(iii) To find the minimum velocity, differentiate \(v\) and set the derivative to zero:

\(\frac{dv}{dt} = 6t - 8\).

Setting \(\frac{dv}{dt} = 0\) gives \(6t - 8 = 0\), so \(t = \frac{8}{6} = \frac{4}{3}\).

Substitute \(t = \frac{4}{3}\) back into the expression for \(v\):

\(v = 3\left(\frac{4}{3}\right)^2 - 8\left(\frac{4}{3}\right) + 4 = 3 \times \frac{16}{9} - \frac{32}{3} + 4\).

\(v = \frac{48}{9} - \frac{96}{9} + \frac{36}{9} = \frac{48 - 96 + 36}{9} = \frac{-12}{9} = -\frac{4}{3}\).