解：（1）$\begin{cases}x + y = 3,① \\ y + z = 5,② \\ z + x = 4,③\end{cases}$
$① - ②，$得$x - z = - 2,④$
$③ + ④，$得$2x = 2，$解得$x = 1.$
将$x = 1$代入$①，$得$1 + y = 3，$解得$y = 2.$
将$y = 2$代入$②，$得$2 + z = 5，$解得$z = 3，$
所以原方程组的解为$\begin{cases}x = 1 \\ y = 2 \\ z = 3\end{cases}.$
