解(1)∵f(x)=asinωx+bcosωx=
sin(ωx+φ)(ω>0),
又f(x)≤f(
)=4恒成立,
∴
=4,即a 2+b 2=16.…①
∵f(x)的最小正周期为π, ∴ω=
=2,
即f(x)=asin2x+bcos2x(ω>0).
又f(x) max=f(
)=4, ∴asin
+bcos
=4,即a+
b=8.…②
由①、②解得a=2,b=2
.
(2)由(1)知f(x)=2sin2x+2
cos2x=4sin(2x+
).
∵0<x<π,
∴
<2x+
<
,
列表如下:
∴函数f(x)的图象如图所示:
(3)∵f(x 1)=f(x 2),由(2)知,当0<x 1<x 2<
时,x 1+x 2=2×
=
,
∴f(x 1+x 2)=f(
)=4
=2
;
当
<x 1<x 2<π时,x 1+x 2=2×
=
,
∴f(x 1+x 2)=f(
)=4sin
=2
;
综上,f(x 1+x 2)=2
.