When lift accelerates up, pseudo force acts downwards, hence it increases apparent weight. W1>mg
When lift is stationary, W2=mg
When lift falls freely, it accelerates with g downwards, causing an upwards pseudo force in the frame of the lift equal to mg. Hence total force is 0. So weight is 0. W3=0
When lift accelerates down, pseudo force acts upwards, hence it decreases apparent weight W4<mg, but also the acceleration is less than g, therefore W4=m(g−a)>0