$Đáp$ $án$ $+$ $Giải$ $thích$ $các$ $bước$ $giải$ $:$

 Ta có `hat{xOm}` = `hat{nOy}` ( đối đỉnh )

$Mà$ `hat{nOy}` $=$ $132^{o}$ $\Longrightarrow$ `hat{xOm}` $=$ $132^{o}$

 $Vì$ `hat{xOn}` $và$ `hat{moy}` $là$ $2$ $góc$ $đối$ $đỉnh$ $nên$ `hat{xOn}` $=$ `hat{moy}`

 $Ta$ $có$ `hat{xOm}` $+$ `hat{nOy}` $+$ `hat{xOn}` $.$ 2 $=$ $360^{o}$

      $\Leftrightarrow$ $132^{o}$ $+$ $132^{o}$ $+$ `hat{xOn}` $.$ $2$ $=$ $360^{o}$

      $\Leftrightarrow$ $264^{o}$ $+$ `hat{xOn}` $.$ 2 $=$ $360^{o}$

      $\Leftrightarrow$ `hat{xOn}` $.$ $2$ $=$ $360^{o}$ $-$ $264^{o}$

      $\Leftrightarrow$ `hat{xOn}` $.$ $2$ $=$ $96^{o}$

      $\Leftrightarrow$ `hat{xOn}` $=$ $96^{o}$ $:$ $2$

      $\Leftrightarrow$ `hat{xOn}` $=$ $48^{o}$

 Ta có `hat{xOn}` $=$ `hat{mOy}` ( $đối$ $đỉnh$ )

$Mà$ `hat{xOn}` $=$ $48^{o}$ $\Longrightarrow$ `hat{mOy}` $=$ $48^{o}$

     $Vậy$ `hat{xOm}` $=$ $132^{o}$ $;$ `hat{xOn}` $=$ $48^{o}$ $;$ `hat{mOy}` $=$ $48^{o}$

@namlebao

    Chúc cậu học giỏi nha, PLS câu trả lời hay nhất.