(i) For particle A on the table, applying Newton's second law:

\(T - \frac{2}{7} \times 1.26 \times g = 1.26 \times a\)

For particle B hanging, applying Newton's second law:

\(0.9 \times g - T = 0.9 \times a\)

Solving these equations simultaneously gives:

\(0.9g - \frac{2}{7} \times 1.26g = (0.9 + 1.26) \times a\)

\(a = 2.5 \text{ m s}^{-2}\)

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

\(T = \frac{2}{7} \times 1.26 \times g + 1.26 \times 2.5\)

\(T = 6.75 \text{ N}\)

(ii) Using \(v^2 = 2ah\) for particle B:

\(v^2 = 2 \times 2.5 \times 0.45\)

\(v = 1.5 \text{ m s}^{-1}\)

(iii) For particle A, using \(v^2 = v_B^2 + 2as\):

\(v^2 = 2.25 + 2 \times \left( -\frac{20}{7} \right) \times 0.03\)

\(v = 1.44 \text{ m s}^{-1}\)