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

For particle A on the table:

\(T - 0.6 = 0.4a\)

For particle B hanging vertically:

\(0.1g - T = 0.1a\)

Adding these two equations to eliminate \(T\):

\(0.1g - 0.6 = 0.5a\)

Solving for \(a\):

\(a = \frac{0.1 \times 9.8 - 0.6}{0.5} = 0.8 \text{ m/s}^2\)

Substitute \(a = 0.8\) back into the first equation to find \(T\):

\(T - 0.6 = 0.4 \times 0.8\)

\(T = 0.92 \text{ N}\)

For part (ii), using \(v = at\) to find the speed of B after 1.5 s:

\(v = 0.8 \times 1.5 = 1.2 \text{ m/s}\)