(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 \(\tan \theta\).