①

Solution: (1) It takes X days for Party A to work alone, Y days for Party B and Z days for Party C, and the whole engineering quantity is 1, then:

[( 1/X)+( 1/Y)]* 6 = 1；

[( 1/Y)+( 1/Z)]* 10 = 1；

[( 1/X)+( 1/Z)]* 5 =(2/3)

Solve the equation:

x = 10；

y = 15；

Z=30

(2) It is required to be alone for no more than 15 days, and only Party A or Party B can finish the work alone;

Let's assume that a gets a yuan, b yuan and c yuan one day, then:

a+B =(8700/6)= 1450；

b+ C =(9500/ 10)= 950；

a+C =(5500/5)= 1 100；

Namely: Party A and Party B get 1.450 yuan every day, Party B and Party C get 950 yuan every day, and Party A and Party C get1.654,38+0.000 yuan every day.

Solve the equation: A=800, B = 650.

However, it takes 10 days for A to work alone, so the amount of money needed for A to work alone is: 800 *10 = 8,000 yuan;

It takes 15 days for Party B to do it alone, so the amount of money needed for Party B to do it alone is: 650* 15=9750 yuan.

Therefore, this project is done by A alone and costs the least.

②

Let the daily expenses of Party A, Party B and Party C be X, Y and Z respectively, and the following equation holds:

6x+6y=8700

10y+ 10z=8000

5x+5z=5500

The solutions are x = 875, y = 575 and z = 225.

Because it takes 65,438+00,65,438+05,30 days for Party A, Party B and Party C to complete independently, Party C should be excluded first.

Party A needs 10 * 875 = 8750 yuan, and Party B needs 15 * 575 = 8625 yuan, so it costs the least to complete it alone in 15 days.