First, resolve the force into horizontal and vertical components. The horizontal component of the force is given by:

\(F_x = 12 \cos 25^{\circ}\)

Using Newton's second law, \(F = ma\), where \(m = 3 \text{ kg}\), we have:

\(12 \cos 25^{\circ} = 3a\)

Solving for \(a\):

\(a = \frac{12 \cos 25^{\circ}}{3} = 4 \cos 25^{\circ} \approx 3.625 \text{ m/s}^2\)

Using the equation of motion \(s = ut + \frac{1}{2}at^2\), where initial velocity \(u = 0\) and \(t = 5 \text{ s}\):

\(s = \frac{1}{2} \times 4 \cos 25^{\circ} \times 5^2\)

Calculating \(s\):

\(s = \frac{1}{2} \times 4 \times 3.625 \times 25 \approx 45.3 \text{ m}\)

(i) Resolve the forces horizontally and vertically:

Horizontal: \(F \cos 70^\circ + 20 - 10 \cos 30^\circ = R \cos 15^\circ\)

Vertical: \(10 \sin 30^\circ - F \sin 70^\circ = R \sin 15^\circ\)

Solving these equations simultaneously gives \(F = 1.90 \text{ N}\) and \(R = 12.4 \text{ N}\).

(ii) Use the equation of motion: \(v^2 = u^2 + 2as\)

\(11.7^2 = 0 + 2a \times 3\)

\(a = 22.815 \text{ m/s}^2\)

Applying Newton's second law: \(R \cos 15^\circ = m \times 22.815\)

\(m = \frac{R \cos 15^\circ}{22.815} = 0.526 \text{ kg}\)

(i) Resolve the forces in the x and y directions:

\(F \cos \theta = 2.5 \times \frac{24}{25} + 2.6 \times \frac{5}{13} = 3.4\)

\(F \sin \theta = 2.6 \times \frac{12}{13} - 2.5 \times \frac{7}{25} = 1.7\)

Using \(F^2 = (F \cos \theta)^2 + (F \sin \theta)^2\), solve for \(F\):

\(F = 3.80 \text{ N}\)

Using \(\tan \theta = \frac{F \sin \theta}{F \cos \theta}\):

\(\tan \theta = 0.5\)

(ii) When the force \(F\) is removed, use Newton's second law:

\(Resultant force = 3.80 N\)

\(3.80 = 0.5a\)

\(a = 7.60 \text{ ms}^{-2}\)

Direction is \(26.6^\circ\) clockwise from the positive x-axis, using the angle found from \(\tan \theta\).

Resolve forces perpendicular to the plane:

\(R = P \sin 35 + 0.6g \cos 35\)

Resolve forces parallel to the plane:

\(F + P \cos 35 = 0.6g \sin 35\)

Using \(F = 0.4R\), substitute into the parallel resolution equation:

\(0.4R + P \cos 35 = 0.6g \sin 35\)

Substitute \(R = \frac{0.6g}{\cos 35 + 0.4 \sin 35}\) into the equation:

\(0.4 \left( \frac{0.6g}{\cos 35 + 0.4 \sin 35} \right) + P \cos 35 = 0.6g \sin 35\)

Solve for \(P\):

\(P = \frac{0.6g \sin 35 - 0.4 \times 0.6g \cos 35}{\cos 35 + 0.4 \sin 35}\)

Calculate \(P\) to get \(P = 1.41 \text{ N}\).

Resolve forces perpendicular to the plane:

\(R + P \sin 30^{\circ} = 0.12g \cos 40^{\circ}\)

Using \(F = \mu R\), where \(\mu = 0.32\):

\(F = 0.32R\)

For the particle about to slip down the plane:

\(P_{\text{min}} \cos 30^{\circ} + F = 0.12g \sin 40^{\circ}\)

For the particle about to slip up the plane:

\(P_{\text{max}} \cos 30^{\circ} - F = 0.12g \sin 40^{\circ}\)

Substitute for \(F\) and solve for \(P\):

\(P \cos 30^{\circ} = 0.12g \sin 40^{\circ} \pm 0.32 (0.12g \cos 40^{\circ} - P \sin 30^{\circ})\)

Solving these equations gives:

\(P_{\text{max}} = 1.04\)

\(P_{\text{min}} = 0.676\)

Thus, the set of possible values for \(P\) is:

\(0.676 \leq P < 1.04\)