(a) To find the greatest height above the ground reached by P, we use the equation for vertical motion:

\(v^2 = u^2 - 2gs\)

where \(v = 0\) m/s (at the highest point), \(u = 5\) m/s, and \(g = 9.8\) m/s².

\(0 = 5^2 - 2 \times 9.8 \times s\)

\(s = \frac{25}{19.6} = 1.25\) m

The height above the ground is \(2.8 + 1.25 = 4.05\) m.

(b) To find the length of time for which P is at a height of more than 3.6 m above the ground, we use the equation:

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

Let \(s = 3.6 - 2.8 = 0.8\) m, \(u = 5\) m/s, \(a = -9.8\) m/s².

\(0.8 = 5t - 4.9t^2\)

Solving the quadratic equation \(4.9t^2 - 5t + 0.8 = 0\), we find \(t = 0.2\) s and \(t = 0.8\) s.

The length of time for which P is above 3.6 m is \(0.8 - 0.2 = 0.6\) s.