(i) For \(t > 8\), the velocity \(v(t) = \frac{1}{2} t^{2/3}\). The acceleration \(a(t)\) is the derivative of \(v(t)\) with respect to \(t\):

\(a(t) = \frac{d}{dt} \left( \frac{1}{2} t^{2/3} \right) = \frac{1}{3} t^{-1/3}\).

At \(t = 8\), the acceleration just before passing \(A\) is \(\frac{1}{4} \text{ m s}^{-2}\) and just after is \(\frac{1}{3} \times 8^{-1/3} = \frac{1}{6} \text{ m s}^{-2}\).

The decrease in acceleration is \(\frac{1}{4} - \frac{1}{6} = \frac{1}{12} \text{ m s}^{-2}\).

(ii) The distance moved from \(t = 0\) to \(t = 8\) is given by:

\(s_1 = \frac{1}{2} \times \frac{1}{4} \times 8^2 = 8 \text{ m}\).

For \(t > 8\), the distance \(s_2\) is:

\(s_2 = \int_{8}^{27} \frac{1}{2} t^{2/3} \, dt = \left[ 0.3 t^{5/3} \right]_{8}^{27}\).

Calculating the definite integral:

\(s_2 = 0.3 \times 27^{5/3} - 0.3 \times 8^{5/3} = 71.3 - 8 = 63.3 \text{ m}\).

Total distance \(s = s_1 + s_2 = 8 + 63.3 = 71.3 \text{ m}\).