解:(1)過點C作CE⊥AB,BE=2,CE=4,
在Rt△BCE中,BC=2

;
(2)∵PQ∥CB,
∴∠QPA=∠B,
∵∠QAP=∠CEB=90°,
∴△APQ∽△EBC,
∴

y=16-2x;
(3)①當∠QCP=90°時,如圖1,
可證△QCD∽△PCE,

,即

解得x=

;
②當∠CQP=90°時,如圖2,可證△CDQ∽△QAP,

∴

,即

解得x
1=7.5,x
2=8(增根,舍去);
③當∠CPQ=90°時,如圖1,
∵PQ∥BC,所以∠PCB=90°,可證△PCE∽△BCE,
∴

,即(2

)
2=2x,
x=10>8,舍去.
綜上,當x=

或x=7.5時,△QCP是直角三角形.
分析:(1)過點C作CE⊥AB,根據(jù)勾股定理即可求出BC的長;
(2)先根據(jù)平行線的性質(zhì)求出△APQ∽△EBC,再由相似三角形的對應邊成比例即可解答;
(3)先根據(jù)題意畫出圖形,由于不明確直角三角形中哪個角是直角,故應分三種情況討論,分別根據(jù)相似三角形的性質(zhì)解答即可.
點評:此題比較復雜,解答此題的關鍵是作出輔助線,構造出相似三角形,利用相似三角形的對應邊成比例解答.在解答(3)時要分類討論,不要漏解.