Because 2 does not belong to (-π/2, π/2) and f(x)=f(π-x) holds for any real number, then f(2)=f(π-2), π-2∈(-π/2, π/2), that is, f(2).

Because 3 does not belong to (-π/2, π/2), in the same way, we can get f (3) = f (π-3) = π-3+sin (π-3) = π-3+sin3.

sin 1=sin57.30，sin 2 = sin 1 14.60 = sin 75.40，sin 3 = sin 17 1 54 ' = sin 7. 10

According to the property of sine function increasing on (0, π/2), so sin 3

Because f (2)-f (1) = π-3+sin2-sin1,π-3 >; 0，sin 2-sin 1 & gt； 0, so f(2)-f( 1)>0, that is, f(2)>f( 1),

F(2)>f(3), f( 1)>F(3), that is, F (2) >: f( 1)>F(3), so b>a>c.