First, use the equation of motion:

\(v = u + at\)

where \(v = 4.5 \text{ m/s}\), \(u = 2.5 \text{ m/s}\), and \(t = 5 \text{ s}\).

Substitute the values:

\(4.5 = 2.5 + a \times 5\)

Solve for \(a\):

\(a = 0.4 \text{ m/s}^2\)

Next, apply Newton's second law:

\(F - 1.5 = 0.2a\)

Substitute \(a = 0.4\):

\(F - 1.5 = 0.2 \times 0.4\)

\(F - 1.5 = 0.08\)

Solve for \(F\):

\(F = 1.58 \text{ N}\)

Given \(\tan \alpha = \frac{3}{4}\), we find \(\cos \alpha = \frac{4}{5}\) and \(\sin \alpha = \frac{3}{5}\).

(i) For Fig. 1, resolve forces vertically:

Normal reaction, \(R_1 = 2g + 12 \sin \alpha\).

Frictional force, \(F = 12 \cos \alpha\).

For equilibrium, \(F \leq \mu R_1\).

\(12 \times \frac{4}{5} \leq \mu (2g + 12 \times \frac{3}{5})\).

\(9.6 \leq \mu (19.6 + 7.2)\).

\(\mu \geq \frac{9.6}{27.2} = \frac{6}{17}\).

(ii) For Fig. 2, resolve forces vertically:

Normal reaction, \(R_2 = 2g - 12 \sin \alpha\).

Frictional force, \(F = 12 \cos \alpha\).

For sliding, \(F > \mu R_2\).

\(12 \times \frac{4}{5} > \mu (19.6 - 7.2)\).

\(9.6 > \mu (12.4)\).

\(\mu < \frac{9.6}{12.4} = \frac{3}{4}\).

(i) The normal reaction force \(R\) on box B is given by:

\(R = (200 + 250)g = 4500 \text{ N}\)

Using the limiting equilibrium condition, \(P = \mu R\), we have:

\(3150 = \mu \times 4500\)

Solving for \(\mu\), we get:

\(\mu = \frac{3150}{4500} = 0.7\)

(ii) For box A not to slide on B, the frictional force must equal the force required to accelerate A:

\(0.2 \times 200g = 200a\)

Solving for \(a\), we find:

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

Thus, \(a \leq 2 \text{ m s}^{-2}\).

(iii) Applying Newton's second law to the system of A and B:

\(P - F = 450a\)

Where \(F\) is the frictional force between A and B:

\(F = 0.2 \times 200g = 400 \text{ N}\)

Thus, the maximum \(P\) is:

\(P_{\text{max}} = 3150 + 450 \times 2 = 4050 \text{ N}\)

(i) The total weight of the boxes is given by:

\(R = 2 \times 400 \times 10 = 8000 \text{ N}\)

The maximum frictional force between the lower box and the ground is:

\(F = \mu R = 0.75 \times 8000 = 6000 \text{ N}\)

Therefore, the boxes will remain at rest if \(P \leq 6000\).

(ii) For no sliding between the boxes, the maximum frictional force between them is:

\(F_{\text{max}} = 0.4 \times 4000 = 1600 \text{ N}\)

Using Newton's second law for the upper box:

\(400a \leq 1600\)

Thus, \(a \leq 4\).

For the entire system, using Newton's second law:

\(P - 6000 = 800 \times a\)

Substituting \(a = 4\):

\(P - 6000 = 800 \times 4\)

\(P = 9200\)

Therefore, the maximum possible value of \(P\) is 9200.

(i) Resolve forces horizontally and vertically:

Horizontal: \(7.5 \cos 60^\circ + 4.5 \cos 20^\circ = F \cos \theta\)

\(7.5 \sin 60^\circ - 4.5 \sin 20^\circ = F \sin \theta\)

Calculate: \(7.5 \cos 60^\circ + 4.5 \cos 20^\circ = 7.97861\)

\(7.5 \sin 60^\circ - 4.5 \sin 20^\circ = 4.95609\)

Use Pythagoras: \(F = \sqrt{7.98^2 + 4.96^2} = 9.39\) N

Use trigonometry: \(\theta = \arctan \left( \frac{4.96}{7.98} \right) = 31.8^\circ\)

(ii) Apply Newton's second law along AB:

\(9.5 \cos 30^\circ - 7.5 \cos 60^\circ - 4.5 \cos 20^\circ = m \times 1.5\)

Calculate: \(m = 0.166\) kg