Let b=-a, so there is f(0)+f(2a)=2f(a)f(-a).

Comparing the above two formulas, we can get 2f(a)f(a)=2f(a)f(-a).

So there is f(a)=0 or f(a)=f(-a).

When f(a)=0, because a is arbitrary, there is f(0)=0, which is inconsistent with the meaning of the question, so there is f(a) which is not equal to 0.

So only f(a)=f(-a)

So this function is an even function.

2. Let a=b=0 and we can get 2f(0)=2f(0)f(0), so we have f(0)= 1.

F(x+T)=f(x) indicates that the function is a periodic function with a period of t.

So f(a+T)=f(a), f(a-T)=f(a).

Let b=T and get f(a+T)+f(a-T)=2f(a)f(T).

So 2f(a)=2f(a)f(T)

So there is f(T)= 1.

Substituting a=b=m gives f (2m)+f (0) = 2f (m) 2 = 0.

So f(2m)=- 1.

Let a=b=2m substitute to get f (4m)+f (0) = 2f (2m) 2 = 2.

So there is f(4m)= 1.

Similarly, F (8m) = 1, F (16m) = 1. ....

For f (4m * 2 k) = 1 [k = 0, 1, 2, 3...]

So there is t = 4m * 2 k [k = 0, 1, 2,3 ...]