(i) Resolve the forces acting at O vertically:

\(W \cos \alpha + 7 \times 0.6 = 8\)

\(W \cos \alpha = 3.8\)

Using the Pythagorean identity, \(W^2 = (W \sin \alpha)^2 + (W \cos \alpha)^2\), we find:

\(W \sin \alpha = 5.6\)

(ii) Using \(W^2 = (W \sin \alpha)^2 + (W \cos \alpha)^2\):

\(W^2 = 5.6^2 + 3.8^2\)

\(W = \sqrt{31.36 + 14.44} = \sqrt{45.8} \approx 6.77\)

Using \(\tan \alpha = \frac{W \sin \alpha}{W \cos \alpha}\):

\(\tan \alpha = \frac{5.6}{3.8}\)

\(\alpha = \arctan\left(\frac{5.6}{3.8}\right) \approx 55.8^\circ\)

To find the tensions in the strings, we resolve the forces horizontally and vertically.

For horizontal equilibrium:

\(0.8T_1 + \frac{12}{13}T_2 = 2.24\)

For vertical equilibrium:

\(0.6T_1 - \frac{5}{13}T_2 = 1.4\)

Solving these two equations simultaneously, we find:

\(T_1 = 2.5 \text{ N}\)

\(T_2 = 0.26 \text{ N}\)

To solve for the tensions in the strings, we resolve the forces acting on P both horizontally and vertically.

1. **Horizontal Resolution:**

The horizontal component of the tension in the string attached to A is \(T_1 \cos 36.9^ ext{o}\) and for the string attached to B is \(T_2 \cos 16.3^ ext{o}\). The horizontal force is 10 N.

Thus, the equation is:

\(T_1 \cos 36.9^ ext{o} + T_2 \cos 16.3^ ext{o} = 10\)

2. **Vertical Resolution:**

The vertical component of the tension in the string attached to A is \(T_1 \sin 36.9^ ext{o}\) and for the string attached to B is \(T_2 \sin 16.3^ ext{o}\). The weight of P is \(0.7g\) N.

Thus, the equation is:

\(T_1 \sin 36.9^ ext{o} - T_2 \sin 16.3^ ext{o} = 0.7g\)

3. **Solving the Equations:**

Using the given values:

\(0.8T_1 + 0.96T_2 = 10\)

\(0.6T_1 - 0.28T_2 = 0.7g\)

Solving these simultaneous equations gives:

\(T_1 = 11.9\) N and \(T_2 = 0.5\) N.

To find the tensions in the strings, we resolve the forces acting on the particle P vertically and horizontally.

Vertically, the sum of the vertical components of the tensions must equal the weight of the particle:

\(T_A \frac{1}{2.6} + T_B \frac{1}{1.25} = 10.5\)

Horizontally, the horizontal components of the tensions must balance each other:

\(T_A \frac{2.4}{2.6} = T_B \frac{0.75}{1.25}\)

Solving these equations simultaneously gives:

\(T_A = 6.5 \text{ N}\)

\(T_B = 10 \text{ N}\)

Given that \(\tan \alpha = \frac{8}{15}\), we can find \(\cos \alpha\) and \(\sin \alpha\) using the identity \(\tan \alpha = \frac{\sin \alpha}{\cos \alpha}\).

Let \(\sin \alpha = \frac{8}{17}\) and \(\cos \alpha = \frac{15}{17}\) based on the Pythagorean identity \(\sin^2 \alpha + \cos^2 \alpha = 1\).

Resolve forces vertically: \(T \cos \alpha = mg\).

Substitute \(\cos \alpha = \frac{15}{17}\) and \(mg = 0.3 \times 9.8 = 2.94\) N:

\(T \times \frac{15}{17} = 2.94\).

\(T = \frac{2.94 \times 17}{15} = 3.4\) N.

Resolve forces horizontally: \(F = T \sin \alpha\).

Substitute \(T = 3.4\) N and \(\sin \alpha = \frac{8}{17}\):

\(F = 3.4 \times \frac{8}{17} = 1.6\) N.