10. (20 points) The function $f(x)$ defined on $\mathbf{R}$ satisfies:
(i) For any real numbers $x, y$,
$$
f(x+y+1)=f(x-y+1)-f(x) f(y) \text {; }
$$
(ii) $f(1)=2$;
(iii) $f(x)$ is an increasing function on the interval $[0,1]$.
(1) Find the values of $f(0), f(-1), f(2)$;
(2) Solve the inequality $f(x)>1$.


10. (1) Let $x=y=0$, we get
$$
\begin{array}{l}
f(1)=f(1)-f(0) f(0) \\
\Rightarrow f(0)=0 ;
\end{array}
$$

Let $x=-1, y=1$, we get
$$
\begin{array}{l}
f(1)=f(-1)-f(-1) f(1)=-f(-1) \\
\Rightarrow f(-1)=-2 ;
\end{array}
$$

Let $x=0, y=1$, we get
$$
f(2)=f(0)-f(0) f(1)=0 \text {. }
$$
(2) First, study a few properties of the function.
(i) Let $x=-1$, we get
$$
f(y)=f(-y)+2 f(y) \Rightarrow f(-y)=-f(y) \text {. }
$$

Thus, the function $f(x)$ is an odd function.
(ii) Let $x=0$, we get
$$
f(1+y)=f(1-y),
$$

which means the graph of $y=f(x)$ is symmetric about the line $x=1$.
(iii) For any $x \in \mathbf{R}$, we have
$$
\begin{array}{l}
f(x)=f(2-x)=-f(x-2) \\
=-f(4-x)=f(x-4),
\end{array}
$$

which means the function $f(x)$ has a period of 4, and the line $x=1$ is a symmetry axis, with the origin $O(0,0)$ as a symmetry center. Therefore, since the function $f(x)$ is increasing in the interval $[-1,1]$, it is also increasing in the interval $[-1,1]$ and decreasing in the interval $[1,3]$.
Now, let's solve the original problem.
Since $f(x)=f(2-x)$, we have
$$
f\left(\frac{1}{3}\right)=f\left(\frac{5}{3}\right) \text {. }
$$

Let $\left\{\begin{array}{l}x+y+1=\frac{5}{3}, \\ x-y+1=1\end{array}\right.$, solving gives $x=\frac{1}{3}, y=\frac{1}{3}$.
Substituting these into the given equation, we get
$$
\begin{array}{l}
f\left(\frac{1}{3}\right)=f\left(\frac{5}{3}\right)=f(1)-f^{2}\left(\frac{1}{3}\right) \\
\Rightarrow f^{2}\left(\frac{1}{3}\right)+f\left(\frac{1}{3}\right)-2=0 .
\end{array}
$$

Since $2=f(1)>f\left(\frac{1}{3}\right)>f(0)=0$, we have $f\left(\frac{1}{3}\right)=1$.
Thus, the solution to the inequality $f(x)>1$ within one period $[-1,3]$ is $\frac{1}{3}1$, and the solution set is
$$
\left\{x \left\lvert\, \frac{1}{3}+4 k<x<\frac{5}{3}+4 k\right., k \in \mathbf{Z}\right\} \text {. }
$$