(i) To find when the particles are travelling in opposite directions, we need to determine when one particle is moving upwards and the other downwards. The time to reach maximum height for P is given by setting the velocity to zero:

\(0 = 12 - gt \Rightarrow t = \frac{12}{g} \approx 1.2 \text{ s}\)

For Q, \(0 = 7 - gt \Rightarrow t = \frac{7}{g} \approx 0.7 \text{ s}\)

Thus, the particles are travelling in opposite directions for \(0.7 < t < 1.2\).

(ii) Given \(3h_P = 8h_Q\), we use the equations for height:

\(h_P = 12t - \frac{1}{2}gt^2\)

\(h_Q = 7t - \frac{1}{2}gt^2\)

Substitute into the given condition:

\(3(12t - \frac{1}{2}gt^2) = 8(7t - \frac{1}{2}gt^2)\)

Simplifying gives:

\(36t - 1.5gt^2 = 56t - 4gt^2\)

\(2.5gt^2 = 20t\)

\(t = \frac{8}{g}\)

Using \(v = u - gt\) for velocities:

\(v_P = 12 - gt\)

\(v_Q = 7 - gt\)

Substitute \(t = \frac{8}{g}\):

\(v_P = 12 - 8 = 4 \text{ m s}^{-1}\)

\(v_Q = 7 - 8 = -1 \text{ m s}^{-1}\)