∵BC is the diameter of⊙ O,

∴∠BPC=90，

∴∠APC=90，

Q is the midpoint of AC,

∴PQ=CQ，

∴∠CPQ=∠PCQ，

OP = OC

∴∠OPC=∠OCP，

∴∠OPC+∠CPQ=∠OCP+∠PCQ=∠BCA=90，

∴OP⊥PQ，

∴ The straight line PQ is tangent to⊙ O; 、

(2) Solution: Connect CE, as shown in the figure.

∫EP is the diameter,

∴∠ ECP = 90, that is ∠ ECO+∠ OCP = 90,

∠∠ Ecology +∠ ECF = 90,

∴∠ECF=∠OCP=∠OPC，

And < f = < f.

∴△FEC∽△FCP，

∴EFCF=CFPF=ECPC，

In Rt△EPC, tan∠OPC=ECPC= 12.

∴EFCF=CFPF= 12，

∴CF=2EF,PF=2CF，

∴PF=4EF，

∴PE=3EF，

That is, 3EF=2×5,

∴EF=253.