(i) The acceleration \(a\) down the plane is given by \(a = g \sin \alpha = g \times \frac{16}{65}\).

Using the equation \(v^2 = u^2 + 2as\), where \(u = 0\), \(v = 8\), and \(s = S\), we have:

\(8^2 = 2 \times \frac{16}{65}g \times S\)

\(64 = \frac{32}{65}gS\)

\(S = \frac{64 \times 65}{32g} = 13\) m

To find the speed when \(s = \frac{1}{2}S = 6.5\) m, use \(v^2 = 2as\):

\(v^2 = 2 \times \frac{16}{65}g \times 6.5\)

\(v = \sqrt{\frac{32}{65}g \times 6.5} = 5.66\) m s^-1

(ii) The time \(T\) to travel distance \(S\) is given by \(v = u + at\), so \(8 = 0 + \frac{16}{65}gT\).

\(T = \frac{8 \times 65}{16g}\)

For \(\frac{1}{2}T\), the distance \(s\) is given by \(s = \frac{1}{2}a(\frac{T}{2})^2\):

\(s = \frac{1}{2} \times \frac{16}{65}g \times (\frac{8 \times 65}{32g})^2\)

\(s = 3.25\) m