1. Let $\mathbf{Z}$ denote the set of all integers. Find all functions $f: \mathbf{Z} \rightarrow \mathbf{Z}$, such that for any integers $a, b$, we have
$$
f(2 a)+2 f(b)=f(f(a+b)) .
$$
1. Denote the equation in the problem as $P(a, b)$.

From $P(0, x)$ we get
$$
f(0)+2 f(x)=f(f(x)) \text {. }
$$

From $P(x, 0)$ we get
$$
f(2 x)+2 f(0)=f(f(x)) \text {. }
$$

From equations (1) and (2) we get
$$
f(2 x)=2 f(x)-f(0) \text {. }
$$

Substituting equations (1) and (3) into the original equation, we get
$$
2 f(a)-f(0)+2 f(b)=f(0)+2 f(a+b) \text {. }
$$

Let $g(x)=f(x)-f(0)$, then $g(0)=0$.
Thus, equation (4) can be written as
$$
g(a+b)=g(a)+g(b) \text {. }
$$

By the Cauchy method, we know that for any integer $n$,
$g(n)=g(1) n$.
Therefore, $f(x)=k x+c(k, c \in \mathbf{Z})$.
Substituting into the original equation, we require that for all integers $a, b$ we have $2 k(a+b)+3 c=k^{2}(a+b)+(k+1) c$
$\Leftrightarrow 2 k=k^{2}$, and $3 c=(k+1) c$
$\Rightarrow k=2$ or 0.
If $k=2$, then $c$ can be any integer;
If $k=0$, then $c=0$.
Thus, the functions that satisfy the conditions are
$$
f(x)=0
$$

or $f(x)=2 x+c$ (where $c$ is any integer).