(1) $f(x)$ is monotonically increasing on $(-\infty, +\infty)$. The proof is as follows:
Let any $x_1, x_2 \in (-\infty, +\infty)$, and $x_1<x_2$, then
since $x_1-x_2<0$, it follows that $f(x_1-x_2)<0$, thus $f(x_1)=f[(x_1-x_2)+x_2]=f(x_1-x_2)+f(x_2)<f(x_2)$
which means $f(x_1)<f(x_2)$,
therefore, $f(x)$ is monotonically increasing on $(-\infty, +\infty)$. $\boxed{\text{6 points}}$
(2) Proof: In $f(x+y)=f(x)+f(y)$, let $x=y=1$, we get $f(2)=f(1)+f(1)=2$.
Let $x=y=0$, we get $f(0)=f(0)+f(0)$, thus $f(0)=0$.
Let $y=-x$, we get $f(-x)+f(x)=0$, i.e., $f(-x)=-f(x)$
Therefore, $2-a_{n+1}=f(2-a_n)$, thus $2-a_{n+1}=f(2)+f(-a_n)$
Therefore, $2-a_{n+1}=2-f(a_n)$, thus $a_{n+1}=f(a_n)$
Next, we use mathematical induction to prove: $\boxed{\text{9 points}}$
① When $n=1$, $0<a_1<1$, the inequality holds;
② Assume when $n=k$ ($k\in\mathbb{N}^*$), the inequality holds, i.e., $0<a_k<1$,
then since $f(x)$ is monotonically increasing on $(-\infty, +\infty)$,
thus $f(0)<a_{k+1}=f(a_k)<f(1)$, therefore $0<a_{k+1}<1$,
which means the inequality also holds when $n=k+1$.
Combining ① and ②, by the principle of mathematical induction, it is known that for any $n\in\mathbb{N}^*$, $0<a_n<1$. $\boxed{\text{14 points}}$