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.