(i) The distance between the particles is given by the difference in their displacements. For particle P moving down the plane, the displacement is given by:

\(s_P = 1.3t + \frac{1}{2}at^2\)

For particle Q moving up the plane, the displacement is:

\(s_Q = -1.3t + \frac{1}{2}at^2\)

The distance between them is:

\(d = s_P - s_Q = (1.3t + \frac{1}{2}at^2) - (-1.3t + \frac{1}{2}at^2) = 2.6t\)

(ii) Given the vertical height difference is 1.6 m when \(t = 2.5\), we use:

\(\sin \alpha = \frac{1.6}{2.6 \times 2.5}\)

The acceleration \(a\) is given by:

\(a = g \sin \alpha = \frac{32}{13} \approx 3.2 \text{ m s}^{-2}\)

(iii) To find the distance travelled by P when Q is at its highest point, set the final velocity of Q to zero:

\(0 = 1.3 - at\)

Solving for \(t\), we get:

\(t = \frac{1.3}{a} \approx 0.528 \text{ s}\)

The distance travelled by P is:

\(d_P = 1.3 \times 0.528 + \frac{1}{2} \times 3.2 \times (0.528)^2 \approx 1.03 \text{ m}\)