(a) To find the magnitude of the force in each of the struts AD and BD, resolve vertically:

\(500 + T \cos 45^{\circ} + T \cos 45^{\circ} - 100g = 0\)

Since \(T_A = T_B = T\), we have:

\(500 + 2T \cos 45^{\circ} = 100g\)

Solving for \(T\):

\(T \cos 45^{\circ} = \frac{100g - 500}{2}\)

\(T = \frac{500}{\sqrt{2}} = 354 \text{ N}\)

(b) To find the value of \(F\) for which the magnitude of the force in the strut AD is zero, resolve horizontally and vertically:

Vertically:

\(T_B \cos 45^{\circ} + 500 - 100g = 0\)

Horizontally:

\(F - T_B \sin 45^{\circ} = 0\)

Solving for \(F\):

\(F = T_B \sin 45^{\circ}\)

Since \(T_B \cos 45^{\circ} = 100g - 500\),

\(T_B = \frac{100g - 500}{\cos 45^{\circ}}\)

\(F = \frac{100g - 500}{\cos 45^{\circ}} \sin 45^{\circ} = 500 \text{ N}\)

Let the tension in the string attached to the wall be \(T\).

Resolve forces horizontally:

\(T \cos \alpha = 4 \sin 60^{\circ}\)

Resolve forces vertically:

\(T \sin \alpha + 4 \cos 60^{\circ} = 0.3g\)

Substitute \(g = 9.8 \, \text{m/s}^2\):

\(T \sin \alpha + 4 \times 0.5 = 0.3 \times 9.8\)

\(T \sin \alpha = 2.94 - 2\)

\(T \sin \alpha = 0.94\)

Using the horizontal equation:

\(T \cos \alpha = 4 \times \frac{\sqrt{3}}{2}\)

\(T \cos \alpha = 2\sqrt{3}\)

Divide the two equations:

\(\tan \alpha = \frac{0.94}{2\sqrt{3}}\)

\(\alpha = \arctan \left( \frac{0.94}{2\sqrt{3}} \right) \approx 76.9^{\circ}\)

Substitute \(\alpha\) back to find \(T\):

\(T = \frac{2\sqrt{3}}{\cos 76.9^{\circ}} \approx 2.05 \text{ N}\)

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

Let the tensions in the strings be \(T_1\) and \(T_2\).

Resolving horizontally:

\(T_1 \cos 60^{\circ} = T_2 \cos 45^{\circ}\)

Resolving vertically:

\(T_1 \sin 60^{\circ} + T_2 \sin 45^{\circ} = 8g\)

Substituting \(g = 9.8 \text{ m/s}^2\), we have:

\(T_1 \cos 60^{\circ} = T_2 \cos 45^{\circ}\)

\(T_1 \sin 60^{\circ} + T_2 \sin 45^{\circ} = 8 \times 9.8\)

Solving these equations simultaneously gives:

\(T_1 = 58.6 \text{ N}\)

\(T_2 = 41.4 \text{ N}\)

An alternative method using Lami's Theorem:

\(\frac{T_1}{\sin 135^{\circ}} = \frac{T_2}{\sin 150^{\circ}} = \frac{80}{\sin 75^{\circ}}\)

Solving for \(T_1\) and \(T_2\) gives the same results:

\(T_1 = 58.6 \text{ N}\)

\(T_2 = 41.4 \text{ N}\)

(a) Resolve horizontally: \(100 - T_U \sin 60 - T_L \sin 60 = 0\)

Resolve vertically: \(T_U \cos 60 - T_L \cos 60 - 5g = 0\)

Perpendicular to \(T_U\): \(T_L \cos 30 + 5g \cos 30 = 100 \cos 60\)

Solve for \(T_L\): \(T_L = 7.74 \text{ N}\)

(b) Resolve horizontally: \(X - T_{up} \sin 60 = 0\)

Resolve vertically: \(T_{up} \cos 60 - 5g = 0\)

Perpendicular to \(T_{up}\): \(5g \cos 30 = X \cos 60\)

Eliminate \(T_{up}\) and solve for \(X\): Least value of \(X = 86.6\)

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

Resolving horizontally:

\(T_P \cos 60^{\circ} = T_R \cos 30^{\circ}\)

Resolving vertically:

\(T_P \sin 60^{\circ} = T_R \sin 30^{\circ} + 0.2g\)

Using \(g = 9.8 \text{ m/s}^2\), solve these equations simultaneously.

From the horizontal resolution:

\(T_P \cdot 0.5 = T_R \cdot \frac{\sqrt{3}}{2}\)

\(T_P = T_R \cdot \sqrt{3}\)

Substitute \(T_P = T_R \cdot \sqrt{3}\) into the vertical resolution:

\(T_R \cdot \sqrt{3} \cdot \frac{\sqrt{3}}{2} = T_R \cdot 0.5 + 0.2 \cdot 9.8\)

\(\frac{3}{2} T_R = 0.5 T_R + 1.96\)

\(T_R = 2 \text{ N}\)

Substitute \(T_R = 2 \text{ N}\) back to find \(T_P\):

\(T_P = 2 \cdot \sqrt{3} \approx 3.46 \text{ N}\)