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

\((M + m)a = (M - m)g\)

where \(M = 0.75\) kg, \(m = 0.25\) kg, and \(g = 9.8\) m/s².

Solving gives:

\(a = \frac{(0.75 - 0.25) \times 9.8}{0.75 + 0.25} = 5\) m/s².

Using the equation \(s = ut + \frac{1}{2}at^2\) with \(u = 0\), \(t = 0.6\) s:

\(s = 0 + \frac{1}{2} \times 5 \times (0.6)^2 = 0.9\) m.

(ii) From the velocity-time graph, the area under the graph from 0 to 0.6 s is:

\(\frac{1}{2} \times 0.6 \times V = 0.9\)

Solving gives:

\(V = 3\) m/s.

Using the equation \(0 = V - g(T - 0.6)\):

\(0 = 3 - 9.8(T - 0.6)\)

Solving gives:

\(T = 0.9\) s.

(iii) The distance travelled upwards is given by:

\(s_{\text{up}} = \frac{1}{2} \times 0.9 \times 3 = 1.35\) m.

The distance travelled downwards is given by:

\(s_{\text{down}} = 0 + \frac{1}{2} \times 9.8 \times (1.6 - 0.9)^2 = 2.45\) m.

Thus, the height \(h\) is:

\(h = s_{\text{down}} - s_{\text{up}} = 2.45 - 1.35 = 1.1\) m.