\(\begin{array}{l}
1)\\
+)\quad \rm C_5H_{10}O_2\\
\begin{array}{|c|c|c|}
\hline
\text{STT}&\text{Đồng phân}&\text{Tên gọi}\\
\hline
1&\rm CH_3-CH_2-CH_2-CH_2-COOH&\rm Pentanoic\\
\hline
2&\rm CH_3-\mathop{C}\limits_{\Big|}\kern-1ptH-CH_2-COOH&\rm 3-metylbutanoic\\
&\rm CH_3\qquad\quad \ \ \ &\\
\hline
3&\rm CH_3-CH_2-\mathop{C}\limits_{\Big|}\kern-1ptH-COOH&\rm 2-metylbutanoic\\
&\rm \qquad \ \ \ CH_3&\\
\hline
&\rm CH_3&\\
4&\rm CH_3-\mathop{C}\limits_{\Big|}^{\Big|}-COOH&\rm 2,2-đimetylpropanoic\\
&\rm CH_3&\\
\hline
\end{array}\\
+)\quad \rm C_4H_{8}O_2\\
\begin{array}{|c|c|c|}
\hline
\text{STT}&\text{Đồng phân}&\text{Tên gọi}\\
\hline
1&\rm CH_3-CH_2-CH_2-COOH&\rm Butanoic\\
\hline
2&\rm CH_3-\mathop{C}\limits_{\Big|}\kern-1ptH-COOH&\rm 2-metylbutanoic\\
&\rm CH_3\ \ &\\
\hline
\end{array}\\
2)\\
\rm X:\quad ROH\\
\rm ROH + Na \longrightarrow RONa + \dfrac12H_2\uparrow\\
0,2\ \rm mol\qquad \qquad \qquad \quad 0,1\ mol\\
\quad M_{ancol} = \dfrac{9,2}{0,2}\\
\Leftrightarrow R + 17 = 46\\
\Leftrightarrow R = 29\quad \rm (C_2H_5)\\
\text{Vậy}\ \rm X:\quad C_2H_5OH\\
3)\\
\rm X:\quad ROH\\
\rm ROH + Na \longrightarrow RONa + \dfrac12H_2\uparrow\\
0,3\ \rm mol\qquad \qquad \qquad \quad 0,15\ mol\\
\quad M_{ancol} = \dfrac{9,2}{0,3}\\
\Leftrightarrow R + 17 = 30,67\\
\Leftrightarrow R = 13,67\\
\text{Vậy không có ancol thỏa mãn đề bài}
\end{array}\)