(i) The normal reaction \(R\) is given by \(R = 0.25g \cos \alpha\). Since \(\sin \alpha = 0.8\), \(\cos \alpha = 0.6\). Thus, \(R = 0.25 \times 9.8 \times 0.6 = 1.5 \text{ N}\).

The frictional force \(F\) is \(F = \mu R = 0.5 \times 1.5 = 0.75 \text{ N}\).

The work done against friction is \(\text{WD} = F \times 8 = 0.75 \times 8 = 6 \text{ J}\).

(ii) Using the work-energy principle, initial kinetic energy at \(P\) is \(\frac{1}{2} \times 0.25 \times 15^2\).

Final kinetic energy at \(Q\) is \(\frac{1}{2} \times 0.25 \times v^2\).

Work done against friction is 6 J, and potential energy gain is \(0.25g \times 8 \times 0.8\).

\(\frac{1}{2} \times 0.25 \times 15^2 = \frac{1}{2} \times 0.25 \times v^2 + 6 + 0.25 \times 9.8 \times 8 \times 0.8\).

Solving for \(v\), \(v = 7 \text{ m s}^{-1}\).

(iii) From \(Q\) to \(R\), kinetic energy lost is potential energy gained.

\(\frac{1}{2} \times 0.25 \times 7^2 = 0.25 \times 9.8 \times H\).

Solving for \(H\), \(H = 2.45 \text{ m}\).

Total height \(h = 6.4 + H = 8.85 \text{ m}\).