(1) Solution: (1) Let the cost of flowers and trees A and B be X yuan and Y yuan respectively.

Judging from the meaning of the question: 2x+3y= 1700.

3x+y= 1500

x=400

Solution: y=300

(2) Set a plant and B plant as (3a+ 10).

Yes: 400a+300 (3a+ 10)

(760-400)a+(540-300)>=2 1600

Solution:160/9 < = a <; =270/ 13

Since a is an integer, ∴a can be 18 or 19 or 20.

So there are three concrete solutions:

(1) planting the first kind of flowers and trees 18 plants and the second kind of flowers and trees 3a+ 10=64 plants;

(2) planting the first kind of flowers and trees 19, and the second kind of flowers and trees 3a+ 10=67;

(3) Plant 20 Class A flowers and trees and 70 Class B flowers and trees with 3A+ 10.