解(2)：根据韦达定理
x1+x2=1
x1x2=p-1
所以
x1²+x2²=(x1+x2)²-2x1x2=1²-2(p-1)=-2p+3
因为
[2+x1(1-x1)][2+x2(1-x2)]
=(2+x1-x1²)(2+x2-x2²)
=2×2+2×x2+2×(-x2²)+x1×2+x1×x2+x1×(-x2²)-x1²×2-x1²×x2-x1²×(-x2²)
=4+2x2-2x2²+2x1+x1x2-x1x2²-2x1²-x1²x2+x1²x2²
=4+2(x1+x2)-2(x1²+x2²)+x1x2-x1x2(x1+x2)+(x1x2)²
=4+2×1-2(-2p+3)+p-1-(p-1)×1+(p-1)²
=4+2+4p-6+p-1-p+1+p²-2p+1
=p²+2p+1
所以
p²+2p+1=9
(p+1)²=9
p+1=±3
p+1=3或p+1=-3
p1=2
p2=-4
根据题意,方程的判别式⊿≥0
⊿=(-1)²-4×1×(p-1)
=1-4p+4
=-4p+5
所以-4p+5≥0,p≤5/4
所以p=-4 (p=2不合题意,应该舍去)