1. Let $f(x)$ be an $n$-degree polynomial, and satisfy
$$
f(k)=\frac{k}{k+1}, k=0,1,2, \cdots, n \text {. Find } f(n+1) \text {. }
$$
1. $k$ is $(k+1) f(i)-k=0, k=0,1$, $2, \cdots, n$; Therefore, the $n+1$ degree polynomial $g(x)=(x+1) f(x)-x$ has $n+1$ roots: $x=0$, $1,2, \cdots, n$. Thus, we can set $g(x)=c x(x-1) \cdots(x-n)$, where $c$ is a constant, i.e.,
$$
\begin{array}{c}
(x+1) f(x)-x=c x(x-1) \cdots(x-n). \\
\text{Let } x=-1, \text{ we get } 1=c(-1)(-2) \\
\cdots(-1-n)=c(-1)^{n+1}(n+1)!.
\end{array}
$$

Thus, $c=\frac{(-1)^{n+1}}{(n+1)!}$.
Therefore, when $x=n+1$, we have
$$
\begin{array}{l}
(n+2) f(n+1)-(n+1) \\
=\frac{(-1)^{n+1}}{(n+1)!}(n+1) n \cdots 1
\end{array}
$$

i.e., $f(n+1)=\frac{(-1)^{n+1}+n+1}{n+2}$.