(a) Resolve the forces vertically and equate to zero:

\(4 \sin 30^\circ + F \sin 60^\circ - 6 = 0\)

Substitute \(\sin 30^\circ = 0.5\) and \(\sin 60^\circ = \frac{\sqrt{3}}{2}\):

\(2 + \frac{F \sqrt{3}}{2} - 6 = 0\)

\(\frac{F \sqrt{3}}{2} = 4\)

\(F = \frac{8}{\sqrt{3}} \approx 4.62\)

(b) Resolve forces either vertically or horizontally:

Vertically: \(F \sin \alpha + 4 \sin 30^\circ - 6 = 0\)

Horizontally: \(F \cos \alpha + 3 - 4 \cos 30^\circ = 0\)

\(F \sin \alpha = 4\)

\(F \cos \alpha = 0.464\)

Using Pythagoras:

\(F^2 = 4^2 + 0.464^2\)

\(F = \sqrt{16 + 0.215296} \approx 4.03\)

Using trigonometry:

\(\alpha = \arctan \left( \frac{4}{0.464} \right) \approx 83.4\)