Solution: (1) Assuming that the renovation speed of A is X, the renovation speed of B is Y, and the total engineering quantity of renovated houses is Z, then Z=6X+6Y=4X+9Y, and X= 1.5Y is calculated, and A is faster than B, so it is more time-saving to choose A. ..
(2) If the weekly renovation cost of Party A is A and Party B is B, then 6A+6B=5.2, 4A+9B=4.8, A=0.6, and B=4/ 15 are calculated. Assume that Party B's renovation period is C, Party A's renovation period is 2/3C, and Party A's renovation cost is A * 2/.