(i) Using the equation of motion, the acceleration is given by:

\(v = u + at\)

\(3.5 = 0 + 10a\)

\(a = 0.35 \text{ m/s}^2\)

Applying Newton's second law horizontally:

\(10\cos(15^{\circ}) - F = 2 \times 0.35\)

\(F = 10\cos(15^{\circ}) - 0.7\)

\(F = 8.96 \text{ N}\)

(ii) Resolving forces vertically, the normal reaction \(R\) is:

\(R = 2g - 10\sin(15^{\circ})\)

The coefficient of friction \(\mu\) is given by:

\(F = \mu R\)

\(\mu = \frac{8.96}{2g - 10\sin(15^{\circ})}\)

\(\mu = 0.515\)