(i) To find the velocity \(v(t)\), integrate the acceleration function \(a(t) = 2t^{2/3}\):

\(v(t) = \int 2t^{2/3} \, dt = 1.2t^{5/3} + C\)

Given that the initial velocity \(v(0) = 2\), we find \(C = 2\). Thus,

\(v(t) = 1.2t^{5/3} + 2\)

Set \(v(t) = 3\) and solve for \(t\):

\(1.2t^{5/3} + 2 = 3\)

\(1.2t^{5/3} = 1\)

\(t^{5/3} = \frac{5}{6}\)

(ii) To find the distance \(s(t)\), integrate the velocity function:

\(s(t) = \int (1.2t^{5/3} + 2) \, dt = 0.45t^{8/3} + 2t + C\)

Given that \(s(0) = 0\), we find \(C = 0\). Thus,

\(s(t) = 0.45t^{8/3} + 2t\)

Substitute \(t^{5/3} = \frac{5}{6}\) to find \(t\):

\(t = \left( \frac{5}{6} \right)^{3/5}\)

Calculate \(s(t)\) using this \(t\):

\(s(t) = 0.45 \left( \frac{5}{6} \right)^{8/5} + 2 \left( \frac{5}{6} \right)^{3/5} \approx 2.13 \text{ m}\)