(i) Resolve the forces vertically. The weight of the ring is given by:

\(0.8g\)

where \(g\) is the acceleration due to gravity. The vertical component of the 7 N force is:

\(7 \sin 45^\circ\)

The normal contact force \(R\) acts vertically upwards. In equilibrium:

\(R + 7 \sin 45^\circ = 0.8g\)

Solving for \(R\):

\(R = 0.8g - 7 \sin 45^\circ\)

Substitute \(g = 9.8\):

\(R = 0.8 \times 9.8 - 7 \times \frac{\sqrt{2}}{2}\)

\(R = 7.84 - 4.95 = 2.89\)

Rounding to 3 significant figures gives 3.05 N.

(ii) In limiting equilibrium, the frictional force \(F\) is given by:

\(F = 7 \cos 45^\circ\)

\(F = 7 \times \frac{\sqrt{2}}{2} = 4.95\)

The coefficient of friction \(\mu\) is given by:

\(\mu = \frac{F}{R} = \frac{4.95}{3.05}\)

\(\mu \approx 1.62\)