(i) Using the equation of motion:

\(v = u + at\)

where \(v = 0.6 \text{ m s}^{-1}\), \(u = 0 \text{ m s}^{-1}\), and \(t = 0.3 \text{ s}\).

Substitute the values:

\(0.6 = 0 + a \times 0.3\)

Solve for \(a\):

\(a = \frac{0.6}{0.3} = 2 \text{ m s}^{-2}\)

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

\(mg - T = ma\)

For particle B:

\(T - (1 - m)g = (1 - m)a\)

Equating the two expressions for tension \(T\):

\(mg - ma = (1 - m)g + (1 - m)a\)

Substitute \(a = 2 \text{ m s}^{-2}\) and \(g = 10 \text{ m s}^{-2}\):

\(10m - 2m = 10 - 10m + 2 - 2m\)

Simplify and solve for \(m\):

\(8m = 12 - 12m\)

\(20m = 12\)

\(m = 0.6\)

Substitute \(m = 0.6\) back to find \(T\):

\(T = 8m = 8 \times 0.6 = 4.8 \text{ N}\)