∵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.