(i) The acceleration \(a\) is the derivative of velocity \(v\) with respect to time \(t\). Thus,

\(a = \frac{dv}{dt} = \frac{d}{dt}(0.75t^2 - 0.0625t^3) = 1.5t - 0.1875t^2\).

Setting \(a = 0\) gives:

\(1.5t - 0.1875t^2 = 0\)

\(0.1875t(8 - t) = 0\)

Thus, \(t = 8\) (since \(t = 0\) is not positive).

(ii) The particle changes direction when its velocity is zero. Setting \(v = 0\):

\(0.75t^2 - 0.0625t^3 = 0\)

\(0.0625t^2(12 - t) = 0\)

Thus, \(t = 12\) (since \(t = 0\) is the starting point).

The distance travelled is the integral of velocity from \(t = 0\) to \(t = 12\):

\(s = \int_0^{12} (0.75t^2 - 0.0625t^3) \, dt\)

\(s = \left[ 0.25t^3 - 0.0625 \frac{t^4}{4} \right]_0^{12}\)

\(s = \left[ 0.25 \times 12^3 - 0.0625 \times \frac{12^4}{4} \right]\)

\(s = 432 - 324 = 108 \text{ m}\)