1. Find all non-zero real-coefficient polynomials $P(x)$, $Q(x)$, satisfying
$$
P(x+Q(y))=Q(x+P(y))
$$

for all real numbers $x$, $y$.
1. When $P(x) \equiv Q(x)(x \in \mathbf{R})$, it clearly satisfies the condition.

Below, assume $P(x)$ and $Q(x)$ are two different polynomials.

If $P(x)$ is a constant, then $Q(x+P(y))$ is a constant. Thus, $Q(x)$ is also a constant.
Hence, $P(x) \equiv Q(x)(x \in \mathbf{R})$, which is a contradiction.
Similarly, if $Q(x)$ is a zero-degree polynomial, then
$$
P(x)=Q(x),
$$
which is a contradiction.
Consider the case where $P(x)$ and $Q(x)$ are both non-constant.
Let $P(x)=a x^{n}+b x^{n-1}+R(x)\left(n \in \mathbf{Z}_{+}\right)$,
$$
Q(x)=c x^{m}+d x^{m-1}+S(x)\left(m \in \mathbf{Z}_{+}\right) \text {, }
$$
where $a$ and $c$ are non-zero real numbers,
$\operatorname{deg} R<n-1$, $\operatorname{deg} S<m-1$.
Consider $P(x+Q(y))$ as a polynomial in $x$, its highest degree term is $a x^{n}$, and the coefficient of $x^{n-1}$ is $a n Q(y)+b$;

Consider $Q(x+P(y))$ as a polynomial in $x$, its highest degree term is $c x^{m}$, and the coefficient of $x^{m-1}$ is $c m P(y)+d$.
Since for all real numbers $x, y$, we have
$$
P(x+Q(y))=Q(x+P(y)) \text {, }
$$
we get $m=n, a=c$, and for any $y \in \mathbf{R}$,
$$
a n Q(y)+b=c m P(y)+d \text {. }
$$

Noting that $m=n, a=c$.
Then for any $y \in \mathbf{R}$,
$$
P(y)=Q(y)+\frac{b-d}{a n} \text {. }
$$

Thus, $\frac{b-d}{a n} \neq 0$ (otherwise $P \equiv Q$, which is a contradiction).
Let $t=\frac{b-d}{a n}$.
From $P(x+Q(y))=Q(x+P(y))$
$\Rightarrow Q(x+Q(y))+t=Q(x+Q(y)+t)$.
In equation (1), replace $x$ with $x+Q(y)$ to get
$Q(x)+t=Q(x+t)$.
Hence, for any $k \in \mathbf{Z}$,
$Q(k t)=Q(0)+k t$.
Noting that $t$ can take any non-zero real number.
Thus, $Q(x)=Q(0)+x(x \in \mathbf{R})$.
Hence, $P(x)=x+Q(0)+t(x \in \mathbf{R})$.
In summary, $P(x) \equiv Q(x)(x \in \mathbf{R})$ or
$P(x)=x+u(x \in \mathbf{R}), Q(x)=x+v(x \in \mathbf{R})$,
where $u, v$ are any real numbers, and $u \neq v$.