1. Find a function $f: \mathbf{R} \rightarrow \mathbf{R}$, such that for any real numbers $x, y$, we have $f(f(x)+y)=2 x+f(f(y)-x)$.
(Czech Republic provided)
Let $y=-f(x)$, then
$$
f(0)=2 x+f(f(-f(x))-x),
$$

i.e., $f(0)-2 x=f(f(-f(x))-x)$,
holds for all $x$.
Thus, for any real number $f(0)-2 x=y$, there exists a real number $z$ such that $y=f(z)$, i.e., the function $f$ is surjective.
Therefore, there exists a real number $a$ such that $f(a)=0$.
Let $x=a$, then the original functional equation becomes
$$
f(y)=2 a+f(f(y)-a),
$$

i.e., $f(y)-a=f(f(y)-a)+a$.
Since $f$ is surjective, for each real number $x$, there exists a real number $y$ such that $x=f(y)-a$. Therefore, $x=f(x)+a$, i.e., $f(x)=x-a$ holds for all real numbers $x$.

Upon verification, for any real number $a$, $f(x)=x-a$ satisfies the original functional equation.