解:(1)根据杠杆平衡条件$F_1l_1 = F_2l_2,$以脚尖为支点,$G = 500\ N,$$l_{G}=0.9\ m + 0.6\ m = 1.5\ m,$$l_{F}=0.9\ m,$则$F\times0.9\ m = 500\ N\times1.5\ m,$解得$F = 300\ N。$
(2)$1\ min = 60\ s,$做一次俯卧撑做功$W = Fs = 300\ N\times0.4\ m = 120\ J,$$1\ min$内做$30$个俯卧撑,总功$W_{总}=120\ J\times30 = 3600\ J,$功率$P=\frac{W_{总}}{t}=\frac{3600\ J}{60\ s}=60\ W。$