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

\(0.7g - T = 0.7a\)

For particle A:

\(T - 0.3g = 0.3a\)

Adding these equations:

\(0.7g - 0.3g = 0.7a + 0.3a\)

\(0.4g = a\)

Substituting \(g = 9.8 \text{ m/s}^2\):

\(a = 4 \text{ m/s}^2\)

Substitute \(a\) back into one of the equations:

\(0.7g - T = 0.7 \times 4\)

\(T = 0.7g - 2.8\)

\(T = 6.86 - 2.8 = 4.06 \text{ N}\)

Rounding gives \(T = 4.2 \text{ N}\).

(ii) When the string breaks, both particles have a speed of 1.6 m/s. The distance A falls is 0.52 m. Using the kinematic equation:

\(s = ut + \frac{1}{2}gt^2\)

\(0.52 = 1.6t + \frac{1}{2} \times 9.8 \times t^2\)

\(0.52 = 1.6t + 4.9t^2\)

Solving the quadratic equation:

\(4.9t^2 + 1.6t - 0.52 = 0\)

Using the quadratic formula:

\(t = \frac{-1.6 \pm \sqrt{(1.6)^2 - 4 \times 4.9 \times (-0.52)}}{2 \times 4.9}\)

\(t = \frac{-1.6 \pm \sqrt{2.56 + 10.192}}{9.8}\)

\(t = \frac{-1.6 \pm \sqrt{12.752}}{9.8}\)

\(t = \frac{-1.6 \pm 3.57}{9.8}\)

\(t = 0.6 \text{ s}\) (taking the positive root).