∴f'(x)=2+sinx>； 0, ∴f(x) is the increasing function on R.

f(π/2+x)=π+2x+sinx

f(π/2-x)=π-2x-sinx

∴f(π/2+x)+f(π/2-x)=2π

And for any function y=f(x): if f(x+a)+f(b-x)=c is satisfied.

Then, this function is symmetric about the center of the point (a/2+b/2, c/2).

The image of f (x) is symmetrical about the center of point (π/2, π).

∵ Sequence {an} is a arithmetic progression with a tolerance of π/8, f (a1)+f (A2)+...+f (A5) = 5π.

∴ points (A 1, F (A 1)), (A2, F (A2), … and (a5, f(a5)), (a4, f(a4)), … are on the function image (a3, f(a3).

Let a3=π/2, then f(a3)=π.

∴a 1=a3-2*π/8=π/4

∴[f(a3)]^2-a 1a5=π^2-π/4*3π/4= 13/ 16π^2

Option d