(i) The frictional force \(F\) required to move the block horizontally is given by \(F = \mu W\), where \(\mu = 1.25\) and \(W\) is the weight of the block. Therefore, \(F = 1.25W\). Since the vertical force required to lift the block is \(W\), and \(F = 1.25W\), the vertical force \(W\) is less than the horizontal force \(F\).

(ii) Using Newton's second law, the net force \(F_{\text{net}} = ma\), where \(m\) is the mass and \(a\) is the acceleration. The mass \(m = \frac{W}{g} = \frac{60}{10} = 6\) kg (assuming \(g = 10\) m/s\(^2\)).

The net force is also given by \(P - F = ma\), where \(F = 1.25 \times 60 = 75\) N is the frictional force.

Thus, \(P - 75 = 6 \times 4\).

Solving for \(P\), we get \(P = 24 + 75 = 99\) N.