解:(1)如图12所以,以A为位似中心,链接AC,AD,分别在AB,AC,AD,AE,上取点B′,C′,D′,E′,使AB'/AB=AC'/AC=AD'/AD=AE'/AE=1/2,连接B′C′,C′D′,D′E′,所得五边形AB′C′D′E′即为所求.(答案不唯一)
(2)如图13所示,以AB上的点O为端点作射线OA,OB,OC,OD,OE,分别在射线OA,OB,OC,OD,OE上取点A′,B′,C′,D′,E′,使OA'/OA=OB'/OB=OC'/OC=OD'/OD=OE'/OE=3,连接A′B′,B′C′,C′D′,D′E′,E′A′,所得五边形A′B′C′D′E′即为所求.(答案不唯一)