(i) To find the time taken, we need to integrate the velocity function to get the displacement function:

\(s = \int v \, dt = \int (4t - \frac{1}{16}t^3) \, dt = 2t^2 - \frac{t^4}{64} + C\)

Given that the distance PQ is 64 m, we set \(s = 64\):

\(2t^2 - \frac{t^4}{64} = 64\)

\(t^4 - 128t^2 + 64^2 = 0\)

\((t^2 - 64)^2 = 0\)

Solving gives \(t^2 = 64\), so \(t = 8\) s.

(ii) The acceleration is the derivative of the velocity:

\(a = \frac{dv}{dt} = \frac{d}{dt}(4t - \frac{1}{16}t^3) = 4 - \frac{3}{16}t^2\)

Set \(a > 0\):

\(4 - \frac{3}{16}t^2 > 0\)

\(\frac{3}{16}t^2 < 4\)

\(t^2 < \frac{64}{3}\)

\(t < \frac{8}{\sqrt{3}} \approx 4.62\)

Thus, the set of values for \(t\) is \(0 < t < \frac{8}{\sqrt{3}}\) or \(0 < t < 4.62\).