(i) Using the equation of motion \(s = ut + \frac{1}{2}at^2\) for the segment PQ:

\(0.8 = 0.6u + \frac{1}{2}a(0.6)^2\)

\(0.8 = 0.6u + 0.18a\)

For the segment QR:

\(0.8 = (u + a \times 0.6) \times 1 + \frac{1}{2}a \times 1^2\)

\(0.8 = 1.6u + 1.28a\)

Solving these equations simultaneously gives:

Deceleration \(a = \frac{2}{3}\) m s^-2

\(u = \frac{23}{15}\) m s^-1

(ii) The normal reaction \(R = mg \cos 3\)

The frictional force \(F = \mu mg \cos 3\)

Using Newton's second law:

\(-mg \sin 3 - \mu mg \cos 3 = m \left( -\frac{2}{3} \right)\)

Solving for \(\mu\):

\(\mu = 0.0144\)