解:(1)$f = 0.08G = 0.08mg = 0.08\times6\times10^{3}\ kg\times10\ N/kg = 4.8\times10^{3}\ N。$
(2)因为汽车匀速行驶,所以$F = f = 4.8\times10^{3}\ N;$$1\ min = 60\ s,$$1\ min$内通过的距离$s = vt = 10\ m/s\times60\ s = 600\ m,$汽车牵引力在$1\ min$内做的功$W = Fs = 4.8\times10^{3}\ N\times600\ m = 2.88\times10^{6}\ J。$
(3)汽车牵引力的功率$P=\frac{W}{t}=\frac{2.88\times10^{6}\ J}{60\ s}=4.8\times10^{4}\ W。$