设函数F(X)对任意X,Y∈R,都有F(X+Y)=F(X)+F(Y),且X>0时,F(X)<0 ,F(1)=-2

1.
f(0) = f(0+0) = f(0) + f(0) = 2f(0) → f(0) = 0
f(-x) = f(x-x-x) = f(x) + f(-x) + f(-x) → f(x) + f(-x) = 0
故f(x)是奇函数
2.
设0≤x≤y≤3
f(y) = f(x+t) = f(x) + f(t) ≤ f(x)，再结合奇偶性可知f(x)在-3≤x≤3上是单调递减函数，所以最大值为f(-3)，最小值为f(3)。
f(3) = f(1+1+1) = 3f(1) = -6
f(-3) = -f(3) = 6