f'(x)=(0-cosx)e^x+( 1-sinx)e^x=( 1-sinx-cosx)e^x=( 1-√2*sin(x+π/4))e^x

When x∈(0, π/2), f' (x)

When x∈(π/2, π), f' (x) >; 0, that is, f(x) monotonically increases.

When x=π/2, f'(x)=0, that is, f(x) takes the minimum value f(π/2)=0.

(2)

First, g (0) = f (0)-1=1-1= 0.

Then for any x>0, if g(x)=0, that is, (1-sinx) ex-sinx-1= 0.

G (-x) = (1+sinx) e (-x)+sinx-1.

The equations on both sides of the equation are multiplied by e X.

g(-x)e^x=(sinx- 1)e^x+ 1+sinx=-g(x)=0

Because e x > 0

So g(-x)=0.

That is to say, except that x=0, the zero point of g(x) is symmetrical about the origin.

So we only need to discuss the number of zeros of g(x) on (0, π) here.

g'(x)=f'(x)-cosx=( 1-√2*sin(x+π/4))e^x-cosx

When x∈(0, π/2), g' (x)

When x∈(π/2, π), g' (x) >; 0, that is, g(x) monotonically increases.

When x=π/2, g'(x)=0, that is, g(x) takes the minimum value g(π/2)=-2.

And g(0)=0, g (π) = eπ-1>; 0

So g(x) has only one zero on (0, π), and there are x 1 and x 1∈(π/2, π).

According to the previous analysis, there are only three zeros (-π, π) on g(x), which are -X 1, 0 and X 1 respectively.

Obviously, the sum of these three zeros is 0.