F(x+2)=f(x), which means that the function is a periodic function with a period of 2.

Then f(20 10)=f(2008)=. . . . =f(0)

Similarly, f (-2011) = f (2011) = f ... = f (1).

Then f (2010)+f (-2011) = f (0)+f (1).

When x belongs to [0,2], f(x)=log2(x+ 1).

f(0)= log2(0+ 1)= log2( 1)= 0

f( 1)= log2( 1+ 1)= log2(2)= 1

So f(0)+f( 1)= 1.

That is, f (2010)+f (-2011) =1.