(1) Let x=y=0, then from the first property f(0-0)+f(0+0)=2f(0)f(0), that is 2f (0) = 2f (0) 2.

Because f(0) is not equal to 0, f (0) =1;

Let x=0, then for any real number y, there is f(0-y)+f(0+y)=2f(0)f(y), that is, f(-y)+f(y)=2f(y), so f(x)

(2) Let x=y have f(x-x)+f(x+x)=2f(x)f(x), that is, f (2x)+ 1 = 2f (x) 2, that is, f (2x) = 2f 2 (x)-65438.

(3) 1, so that y=a has f(x-a)+f(x+a)=2f(x)f(a), because f(a)=0, then f(x-a)+f(x+a)=0, and x+a is used.

2. Because f(x)+f(x+2a)=0, there is f (x) =-f (x+2a) (formula 1), and x in this formula is replaced by x+2a.

Then f(x+2a)=-f(x+2a+2a) is f (x+2a) =-f (x+4a) (Equation 2);

F(x)=-F(x+2a)=-(-f(x+4a)) can be obtained from formula 1 and formula 2, that is, f(x)=f(x+4a), so f (x) is a periodic function and 4a is a period.