1)

Because: AB=AC=2, ∠ A = 90.

So: ∠ B = ∠ C = 45 ......................... (1)

Connect AP, and point P is the midpoint of hypotenuse BC of isosceles right triangle.

So: AP=BP=CP=BC/2, AP⊥BC.

∠ economically active population =∠FAP=45

According to the triangle exterior angle theorem:

∠EAP+∠EPA=∠BEP=45 +∠EPA

∠CPF =∠APC-∠APF = 90-(∠EPF-∠EPA)= 90-45+∠EPA = 45+∠EPA

So: ∠ BEP = ∠ CPF...............(2)

Therefore, from (1) and (2), we can know △BPE∽△CFP.

2)

BE=x，AB=AC=2，BC=2√2，AP=BP=CP=√2

According to the cosine theorem are:

PE^2=BE^2+BP^2-2BE*BPcos45

pe^2=x^2+2-2√2x*(√2/2)=x^2-2x+2

PE=√(x^2-2x+2)

According to the triangle similarity of 1):

BE/CP=PE/FP

x/√2=√(x^2-2x+2)/PF

So: pf = √ (2x 2-4x+4)/x

So the area of the triangle PEF:

S=y=PE*PF*(sin45 )/2

=√(x^2-2x+2)*[√2*√(x^2-2x+2)/x]*(√2/4)

=(x^2-2x+2)/(2x)

So:

y=x/2+ 1/x- 1，0 & lt； x & lt2

3)

∠EFP=60，∠EPF=45，∠FEP=75

According to sine theorem:

PE/sin60 =PF/sin75 =2R

√(x^2-2x+2)/=√2*√(x^2-2x+2 sin 60)/(xsin 75)

So:

√2sin60 =xsin75

x =(√6/2)/(sin 45 cos 30+cos 45 sin 30)

x=(√6/2) /(√6/4+√2/4)

x=2√3/(√3+ 1)

x=√3*(√3- 1)

X=3-√3 belongs to (0,2)

So: x=3-√3.