解:(1)∵点A(﹣1,0),B(1,0),动点P(x,y),
∴ ,
,
∵PA与PB的斜率之积为3,
∴ ,x≠±1,
∴ .
(2)①∵∠CFB=α,∠CBF=β,β为锐角,
∵tanα= ,tanβ=
,
,
∴tan2β= =
=
=tanα.
②由题意C(x1,y1),y1>0,D(x2,y2),y2<0,
联立 ,得2y2+6by﹣9b2+9=0,
则△=36b2﹣4×2(﹣9b2+9)>0,
∴ , y1+y2=﹣3b,
,
∴b2>1.故y2﹣y1=﹣3b, ,
∴b2>1,故 ,
设∠DFB=γ,∠DBF=θ,
∵ ,tan
,
,
∴tan2θ= =
=﹣
=tanγ,
∵2θ∈(0,π),γ∈(0,π),
∴γ=2θ,即∠DFB=2∠DBF,
∵α,2β∈(0,π),
∴由(2)①得α=2β,即∠CFB=2∠CBF,
又∠DFB=2∠DBF,
∴∠FCB与∠FDB互补,即∠FCB+∠FDB=π,
∴2π﹣3∠CBF﹣3∠DBF=π,则 ,
由到角公式,得 =
, ∴
=
,
即 ,
∴4b+4=﹣ ,解得b=﹣
,满足b2>1,
∴b=﹣ .
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