For part (i), apply Newton's second law to both particles. For particle A on the inclined plane:

\(T - 0.26g \sin \alpha = 0.26a\)

For particle B hanging freely:

\(0.52g - T = 0.52a\)

Solving these equations simultaneously:

\(T = 0.26g \sin \alpha + 0.26a\)

\(0.52g - (0.26g \sin \alpha + 0.26a) = 0.52a\)

\(0.52g - 0.26g \sin \alpha = 0.78a\)

\(a = \frac{0.52g - 0.26g \sin \alpha}{0.78}\)

Substitute \(g = 9.8\), \(\sin \alpha = \frac{16}{65}\):

\(a = \frac{0.52 \times 9.8 - 0.26 \times 9.8 \times \frac{16}{65}}{0.78} = 5.85 \text{ m/s}^2\)

Substitute \(a\) back to find \(T\):

\(T = 0.26 \times 9.8 \times \frac{16}{65} + 0.26 \times 5.85 = 2.16 \text{ N}\)

For part (ii), use the equation \(v^2 = u^2 + 2as\) for particle B:

\(v^2 = 2 \times 5.85 \times 0.6\)

\(v = \sqrt{2 \times 5.85 \times 0.6} = 2.65 \text{ m/s}\)

For particle A, use \(v^2 = u^2 - 2g \sin \alpha \times s\):

\(0 = 2.65^2 - 2 \times 9.8 \times \frac{16}{65} \times s\)

\(s = \frac{2.65^2}{2 \times 9.8 \times \frac{16}{65}} = 1.425 \text{ m}\)

Distance AP when A comes to rest:

\(2.5 - 0.6 - 1.425 = 0.475 \text{ m}\)