(i) Apply Newton's second law to the system:

For block A: \(T - 4g \times \frac{7}{25} = 4a\)

For block B: \(36 - T - 5g \times \frac{7}{25} = 5a\)

For the system: \(36 - 5g \times \frac{7}{25} - 4g \times \frac{7}{25} = 9a\)

Solving these equations gives \(a = 1.2 \text{ ms}^{-2}\) and \(T = 16 \text{ N}\).

(ii) Use the equation of motion:

\(0.65 = 1 \times t + \frac{1}{2} \times 1.2 \times t^2\)

Solving for \(t\) gives \(t = 0.5 \text{ s}\).

(i) For equilibrium, resolve forces along the plane:

\(100 \cos \theta = 8g \sin 30\)

\(100 \cos \theta = 8 \times 9.8 \times 0.5\)

\(100 \cos \theta = 39.2\)

\(\cos \theta = \frac{39.2}{100}\)

\(\theta = \cos^{-1}(0.392) \approx 66.4^{\circ}\)

Resolve forces perpendicular to the plane to find the normal reaction \(R\):

\(R = 8g \cos 30 + 100 \sin \theta\)

\(R = 8 \times 9.8 \times \frac{\sqrt{3}}{2} + 100 \sin 66.4\)

\(R = 67.84 + 91.8\)

\(R = 161\, \text{N}\)

(ii) Given \(\theta = 30\), find the acceleration \(a\):

Apply Newton's second law parallel to the plane:

\(100 \cos 30 - 8g \sin 30 = 8a\)

\(100 \times \frac{\sqrt{3}}{2} - 8 \times 9.8 \times 0.5 = 8a\)

\(86.6 - 39.2 = 8a\)

\(47.4 = 8a\)

\(a = \frac{47.4}{8} = 5.925\)

Thus, \(a \approx 5.83\, \text{m/s}^2\).

(i) Using the equation of motion \(v^2 = u^2 + 2as\), where \(v = 1.5\) m s\(^{-1}\), \(u = 2.5\) m s\(^{-1}\), and \(s = 4\) m, we have:

\(1.5^2 = 2.5^2 + 2a \times 4\)

\(2.25 = 6.25 + 8a\)

\(8a = 2.25 - 6.25\)

\(8a = -4\)

\(a = -0.5\) m s\(^{-2}\)

The deceleration is 0.5 m s\(^{-2}\).

(ii) Using Newton's second law, the acceleration \(a = -g \sin \alpha\).

Given \(a = -0.5\) m s\(^{-2}\), we have:

\(-0.5 = -g \sin \alpha\)

\(\sin \alpha = \frac{0.5}{g}\)

Assuming \(g = 9.8\) m s\(^{-2}\),

\(\sin \alpha = \frac{0.5}{9.8}\)

\(\alpha = \sin^{-1}(0.051)\)

\(\alpha \approx 2.9^\circ\)

(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:

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

To solve for \(u\) and \(a\), we use the equations of motion. For the segment AB, we have:

\(s = ut + \frac{1}{2}at^2\)

Substituting the known values for AB:

\(1.76 = 0.8u + \frac{1}{2} \times a \times (0.8)^2\)

\(1.76 = 0.8u + 0.32a\)

For the segment AC, the total time is 1.4 s:

\(1.76 + 2.16 = (0.8 + 0.6)u + \frac{1}{2}(0.8 + 0.6)^2a\)

\(3.92 = 1.4u + 0.98a\)

Solving these two equations simultaneously:

\(1.76 = 0.8u + 0.32a\)

\(3.92 = 1.4u + 0.98a\)

We find \(u = 1.4\) and \(a = 2\).

For \(\theta\), we use the relation \(a = g \sin \theta\):

\(2 = 10 \sin \theta\)

\(\sin \theta = 0.2\)

\(\theta = 11.5^\circ\)