(1)令a=b=0则f(0)=f(0)+f(0)∴f(0)=0令a=1则f(b)=f(1)+f(b)∴f(1)=0(2)f(36)=f(6×6)=f(6)+f(6)=2f(6)=2f(2×3)=2[f(2)+f(3)]=2(p+q)