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

\(2.4g - T = 2.4a\)

For particle B:

\(T - 1.2g = 1.2a\)

Adding these equations:

\(2.4g - 1.2g = (2.4 + 1.2)a\)

\(1.2g = 3.6a\)

\(a = \frac{1.2g}{3.6} = \frac{10}{3}\) m/s²

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

\(T = 1.2g + 1.2 \times \frac{10}{3}\)

\(T = 12 + 4 = 16\) N

(b) Using energy conservation for particle B:

Initial potential energy = \(1.2g \times 1.5\)

Final potential energy = \(1.2g \times h\)

Using \(v^2 = u^2 + 2as\) with \(v^2 = 2 \times \frac{10}{3} \times 2.1\)

\(v^2 = 14\)

Using \(v^2 = u^2 + 2gh\) with \(u = 0\):

\(0 = 14 - 2g \times s\)

\(s = 4.3\) m

Greatest height = initial height + additional height = \(1.5 + 2.1 + 0.7 = 4.3\) m