解:(1)∵f(x)=a[cos^2(x)+sinxcosx]+b=a[(1+cos2x)/2+(1/2)(2sinxcosx)]+b=a[(1/2)sin2x+(1/2)cos2x+1/2]+b=a[(1/2)(sin2x+cos2x)]+(a+2b)/2=(√2a/2)sin(2x+π/4)+(a+2b)/2∵a>0∴当f(x)单调递增时,2kπ-π/2=...