分析:首先把x2-y2变为x2+xy-xy-y2即x2+xy-(xy+y2)及把x2+2xy+y2变为x2+xy+xy+y2,然后把已知等式的结果代入即可求出x2-y2和x2+2xy+y2的值.
解答:解:(1)∵x2-y2=x2+xy-xy-y2=x2+xy-(xy+y2)
而x2+xy=2,xy+y2=-1,
∴x2-y2=2-(-1)=3;
(2)∵x2+2xy+y2=x2+xy+xy+y2,
而x2+xy=2,xy+y2=1,
∴x2+2xy+y2
=x2+xy+xy+y2
=2-1
=1;
故答案为:3,1.
点评:此题主要考查代数式求值求值,关键是给所求代数式添项,再利用整体代入的数学思想求值.