First, resolve the forces perpendicular to the plane to find the normal reaction \(R\):

\(R = 13g \cos 22.6 = 13g \times \frac{12}{13} = 120 \text{ N}\)

Next, calculate the frictional force \(F\) using \(F = \mu R\):

\(F = 0.3 \times 120 = 36 \text{ N}\)

Apply Newton's second law parallel to the plane (since the particle moves at constant speed, acceleration \(a = 0\)):

\(T = F + 13g \sin 22.6 = 36 + 13g \times \frac{5}{13} = 86 \text{ N}\)

The work done \(WD\) by the force \(T\) is given by:

\(WD = T \times d = 86 \times 2.5 = 215 \text{ J}\)

(i) Resolve forces perpendicular to the plane: \(R = 13 \cos 67.4 = 13 \times \frac{5}{13} = 5\). Resolve forces parallel to the plane: \(F + 13 \sin 67.4 = F + 13 \times \frac{12}{13} = 20\). Using \(F = \mu R\), we have \(\mu = \frac{8}{5} = 1.6\).

(ii) Resolve forces parallel to the plane: \(13 \sin 67.4 - F = 1.3a\). Using \(F = \mu R = 8\), we find \(a = 3.08 \text{ m/s}^2\).

(iii) Use \(s = ut + \frac{1}{2}at^2\) with \(u = 0\) and \(a = \frac{40}{13}\) to find the distance moved in the first 2 seconds: \(s = 0 + 0.5 \times \frac{40}{13} \times 2^2 = \frac{80}{13} = 6.15\). Work done \(\text{WD} = F \times d = 8 \times 6.15 = 49.2 \text{ J}\).

(i) Using the work-energy principle, the change in kinetic energy is equal to the work done against resistance:

\(\frac{1}{2} \times 1.2 \times 7.5^2 - \frac{1}{2} \times 1.2 \times v^2 = 25\)

Solving for \(v\):

\(\frac{1}{2} \times 1.2 \times 7.5^2 = 33.75\)

\(33.75 - 25 = \frac{1}{2} \times 1.2 \times v^2\)

\(8.75 = 0.6v^2\)

\(v^2 = \frac{8.75}{0.6}\)

\(v = \sqrt{14.5833} \approx 3.82 \text{ m s}^{-1}\)

(ii) Considering the inclined plane, the potential energy change is \(1.2g \sin 30 \times d\).

Using the work-energy principle:

\(\frac{1}{2} \times 1.2 \times 7.5^2 - 25 + 1.2g \sin 30 \times d = \frac{1}{2} \times 1.2 \times 9^2\)

\(33.75 - 25 + 1.2 \times 9.81 \times 0.5 \times d = 48.6\)

\(8.75 + 5.886d = 48.6\)

\(5.886d = 39.85\)

\(d = \frac{39.85}{5.886} \approx 6.64 \text{ m}\)

First, calculate the kinetic energy gained by the van:

\(\text{KE gained} = \frac{1}{2} \times 2500 \times (30^2 - 20^2) = 625,000 \text{ J}\)

Next, calculate the potential energy lost as the van descends the hill:

\(\text{PE lost} = 2500 \times 9.8 \times 400 \times \sin 4^{\circ} \approx 697,564.7 \text{ J}\)

Using the work-energy principle, the work done by the engine plus the potential energy lost equals the work done against resistance plus the kinetic energy gained:

\(\text{WD by engine} + \text{PE lost} = 600 \times 400 + 625,000\)

Solving for the work done by the engine:

\(\text{WD by engine} = 600 \times 400 + 625,000 - 697,564.7\)

\(\text{WD by engine} = 167,000 \text{ J}\)