解(1)∵f(x)=asinωx+bcosωx= sin(ωx+φ)(ω>0),
又f(x)≤f( )=4恒成立,
∴ =4,即a2+b2=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(x1)=f(x2),由(2)知,当0<x1<x2< 时,x1+x2=2× = ,
∴f(x1+x2)=f( )=4 =2 ;
当 <x1<x2<π时,x1+x2=2× = ,
∴f(x1+x2)=f( )=4sin =2 ;
综上,f(x1+x2)=2 .
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