Solution: (1) As shown in the figure, if the center of the arc FG is Q, the point crossing Q is the vertical line of CD, the vertical foot is point T, the intersection of MN or its extension line and S, and PQ is connected, then point N is the vertical line of TQ and the vertical foot is W. 。

In Rt△NWS, because NW=2, ∠SNW=θ,

So ns = 2 cos θ.

Because MN and arc FG are tangent to point P, PQ⊥MN,

At Rt△QPS, since PQ= 1, ∠PQS=θ,

So QS = 1cosθ, QT? QS=2？ 1cosθ，

(1) if m is on the line TD, that is, s is on the line TG, then TS=QT-QS,

In Rt△STM, ms = tssinθ = Qt? QSsinθ，

So MN=NS+MS=NS+QT? QSsinθ。

② If m is on CT, that is, S is on the extension line of GT, then TS=QS-QT,

In Rt△STM, ms = tssinθ = QS? QTsinθ，

So MN=NS-MS=NS? QS？ QTsinθ=NS+QT？ QSsinθ。

f(θ)=MN=NS+QT？ QSsinθ=2cosθ+(2sinθ？ 1sinθcosθ)=2(sinθ+cosθ)？ 1sinθcosθ？ (0