1. Set $A=\{x, y\}$ and $B=\left\{1, \log _{3}(x+2)\right\}$ have exactly one common element $1+x$, then $A \cup B=$
1. $\left\{0,1, \log _{3} 2\right\}$ or $\{-1,0,1\}$.

Since $1+x \neq x$, we have $1+x=y$. By $\log _{3}(x+2) \neq 1$ we know $x \neq 1$, so it can only be $y=1+x=1$ or $y=1+x=\log _{3}(x+2)$, thus $(x, y)=(1,0)$ or $(-1,0),(A, B)=\left(\{0,1\},\left\{1, \log _{3} 2\right\}\right)$ or $(\{-1,0\},\{0,1\})$. Therefore, $A \cup B=\left\{0,1, \log _{3} 2\right\}$ or $\{-1,0,1\}$.