(i) Using Newton's second law for both particles, we have:

\(T - 0.2 \times 3 = 0.3a \quad \text{and} \quad 7 - T = 0.7a\)

Solving these equations gives the acceleration:

\(a = 6.4 \text{ m/s}^2\)

Using the equation \(v = u + at\) to find the speed when the string breaks:

\(v = 0 + 6.4 \times 0.25 = 1.6 \text{ m/s}\)

Using \(s = ut + \frac{1}{2}at^2\) to find the distance moved before the break:

\(s = 0 + \frac{1}{2} \times 6.4 \times (0.25)^2 = 0.2 \text{ m}\)

Using \(v^2 = u^2 + 2gs\) to find the speed when B hits the floor:

\(v^2 = 1.6^2 + 2 \times 9.8 \times (0.5 - 0.2)\)

Solving gives:

\(v = 2.93 \text{ m/s}\)

(ii) For finding the distance travelled by A after the break from \(v^2 = u^2 + 2as\):

Distance travelled after break:

\(= (0 - 1.6^2) \div (2 \times -2) = 0.64 \text{ m}\)

Total distance travelled:

\(= 0.2 + 0.64 = 0.84 \text{ m}\)