To solve this problem, we apply Newton's second law to both masses.

For block B on the table, the frictional force is given by:

\(F = 0.5 imes 0.6g\)

For particle A hanging vertically, the forces are:

\(0.4g - T = 0.4a\)

For block B on the table, the forces are:

\(T - F = 0.6a\)

Substituting the expression for frictional force:

\(T - 0.5 imes 0.6g = 0.6a\)

We solve these two equations simultaneously:

1. \(0.4g - T = 0.4a\)

2. \(T - 0.3g = 0.6a\)

Adding the two equations to eliminate T:

\(0.4g - 0.3g = 0.4a + 0.6a\)

\(0.1g = a\)

Thus, the acceleration \(a = 1 \text{ m/s}^2\).

Substitute \(a = 1\) into equation 1:

\(0.4g - T = 0.4 \times 1\)

\(T = 0.4g - 0.4\)

\(T = 3.6 \text{ N}\)

For part (ii), using the equation of motion:

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

Where \(s = 3 \text{ m}\), \(u = 0\), \(a = 1 \text{ m/s}^2\):

\(3 = 0 + \frac{1}{2} \times 1 \times t^2\)

\(t^2 = 6\)

\(t = \sqrt{6} \approx 2.45 \text{ s}\)