1.解:(1)3(xy-2z)+(-xy+3z)=3xy-6z-xy+3z=2xy-3z.
(2)-4(pq+pr)+(4pq+pr)=-4pq-4pq+4pq+pr=-3pr.
(3)(2x-3y)-(5x-y)=2x-3y-5x+y=-3x-2y.
(4)-5(x-2y+1)-(1-3x+4y)=-5x+10y-5-1+3x-4y=-2x+6y-6.
(5)(2a²b-5ab)-2(-ab-a²b)=2a²b-5ab+2ab+2a²b=4a²b-3ab.
(6)1-3(x-1/2 y²)+(-x+1/2 y^2)=1-3x+3/2 y^2-x+ 1/2 y^2=2y^2-4x+1.
2.解:有道理.因为a²+a(a+b)-2a²-ab=a²+a²+ab-2a²-ab=2a²+ab-2a²-ab=0,所以无论a,b区何值,代数式a²+a(a+b)-2a²-ab的值都为0.所以小刚的说法有道理.