(1)f(x)=2sinxcosx+2(cosx)^2=sin2x+cos2x+1=√2sin(2x+π/4)+1
2kπ-π/2<=2x+π/4<=2kπ+π/2,kπ-3π/8<=x<=kπ+π/8.
所以,f(x)的单调递增区间是[kπ-3π/8,kπ+π/8].
(2)a-b=(2sinx-cosx,-cosx),a+b=(2sinx+cosx,3cosx).
若a-b与c共线,则(2sinx-cosx)：2=-cosx：1,2sinx=-cosx.
(a+b)*c=4sinx+2cosx+3cosx=-2cosx+5cosx=3cosx.