(i) Applying Newton's second law to particle A:

\(0.35g - T = 0.35a\)

For particle B:

\(T - 0.15g = 0.15a\)

Adding the two equations:

\((0.35 - 0.15)g = (0.35 + 0.15)a\)

Solving for acceleration \(a\):

\(0.2g = 0.5a\)

\(a = \frac{0.2g}{0.5} = 4\, \text{m/s}^2\)

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

\(T = 0.15g + 0.15 \times 4 = 2.1\, \text{N}\)

(ii) Using the equation \(v_1^2 = 0 + 2a \times 1.6\):

\(v_1^2 = 8 \times 1.6 = 12.8\)

Using the formula for greatest height \(H\):

\(H = 1.6 + \frac{0 - v_1^2}{-2g}\)

\(H = 1.6 + \frac{-12.8}{-20} = 2.24\, \text{m}\)