Bulb resistance RL=U2, p = (8V) 28W = 8ω;

When IL=0.5A, then according to P=UI=I2R:

The power is p = i2r = (0.5a) 2× 8Ω = 2w.

Power consumption w = pt = 2w× 60s = 120J.

So, the answer is: 2; 120.