(i) Resolve forces perpendicular to the plane:

\(R = 0.2g \cos 30^{\circ} - T \sin 15^{\circ}\)

Frictional force \(F = \mu R = 0.3 \times (0.2g \cos 30^{\circ} - T \sin 15^{\circ})\)

Resolve forces parallel to the plane:

\(T \cos 15^{\circ} + 0.3 \times (0.2g \cos 30^{\circ} - T \sin 15^{\circ}) = 0.2g \sin 30^{\circ}\)

Solve for \(T\):

\(T = 0.541\) N

(ii) Work done against friction:

\(0.3 \times 0.2g \cos 30^{\circ} \times 3 = 1.5588\) J

Work done against 0.25 N force:

\(0.25 \times 3 = 0.75\) J

Potential energy loss:

\(0.2g \times 3 \sin 30^{\circ} = 3\) J

Using energy conservation:

\(\frac{1}{2} (0.2) v^2 = 3 - 1.5588 - 0.75\)

Solve for \(v\):

\(Speed = 2.63 m/s\)