Solution: (1)∵ If only 1 car is sold in that month, the purchase price of this car is 270,000 yuan. For every 1 car sold, the purchase price of all cars sold will be reduced by 0. 1 ten thousand yuan.

If the company sells three cars in the same month, the purchase price of each car is: 27-0. 1×2=26.8.

So the answer is: 26.8;

(2) Suppose X cars need to be sold,

According to the meaning of the question, the sales profit of each car is:

28-[27-0.1(x-1)] = (0.1x+0.9) (ten thousand yuan),

When 0≤x≤ 10,

According to the question, did you get an X? (0. 1x+0.9)+0.5x= 12，

Finishing, x2+ 14x- 120=0,

Solving this equation, we get x 1=-20 (irrelevant, omitted), x2=6,

When x > 10,

According to the question, did you get an X? (0. 1x+0.9)+x= 12，

Finishing, x2+ 19x- 120=0,

Solving this equation, we get x 1=-24 (irrelevant, omitted), x2=5,

Because 5 < 10, discard x2=5.

A: Six cars need to be sold.