1)Số đó có dạng $\overline{ab}(a>b)$
Theo bài ra ta có:
$\left\{\begin{array}{l} a=2b-1\\ \overline{ab}-\overline{ba}=18\end{array} \right.\\ <=>\left\{\begin{array}{l} a=2b-1\\ 10a+b-10b-a=18\end{array} \right.\\ <=>\left\{\begin{array}{l} a=2b-1\\ 9a-9b=18\end{array} \right.\\ <=>\left\{\begin{array}{l} a=2b-1\\ a-b=2\end{array} \right.\\ <=>\left\{\begin{array}{l} a=2b-1\\ 2b-1-b=2\end{array} \right.\\ <=>\left\{\begin{array}{l} a=5\\ b=3\end{array} \right.$
$=>$Số đó là $53$
2)Số đó có dạng $\overline{ab},$
Theo bài ra ta có:
$\left\{\begin{array}{l} a+b=9\\\left[\begin{array}{l} a=\dfrac{1}{8}b\\a=8b\end{array} \right.\\\end{array} \right.\\ <=>\left[\begin{array}{l} \left\{\begin{array}{l} a+b=9\\a=\dfrac{1}{8}b\end{array} \right.\\\left\{\begin{array}{l} a+b=9\\a=8b\end{array} \right.\\\end{array} \right.\\ <=>\left[\begin{array}{l} \left\{\begin{array}{l} \dfrac{1}{8}+b=9\\a=\dfrac{1}{8}b\end{array} \right.\\\left\{\begin{array}{l} 8b+b=9\\a=8b\end{array} \right.\\\end{array} \right.\\ <=>\left[\begin{array}{l} \left\{\begin{array}{l} b=8\\a=1\end{array} \right.\\\left\{\begin{array}{l} b=1\\a=8\end{array} \right.\\\end{array} \right.$
=>Vậy số đó là $81$ hoặc $18$
