$a)$
$2H_2O+2Na\rightarrow 2NaOH+H_2$
$x 0,5x$
$2C_2H_5OH+2Na\rightarrow 2C_2H_5ONa+H_2$
$y 0,5y$
$\Rightarrow \begin{cases}0,5x+0,5y=\frac{6,72}{22,4} \\18x+46y=22 \end{cases}\Rightarrow \begin{cases}x=0,2 mol \\ y=0,4 mol \end{cases}\Rightarrow \begin{cases}V_{H_2O}=\frac{0,2.18}{1}=3,6 ml \\V_{C_2H_5OH}=\frac{0,4.46}{0,8}=23 ml \end{cases}$
$\Rightarrow ^oRượu=\frac{23.100}{3,6+23}=86,47^o$
$b)$
$50^o\Rightarrow V_{H_2O}=V_{Rượu}=V ml\Rightarrow \begin{cases}n_{H_2O}=\frac{1.V}{18} \\n_{Rượu}=\frac{0,8.V}{46} \end{cases}$
$\Rightarrow \frac{V}{18}+\frac{0,8V}{46}=2.\frac{6,72}{22,4}=0,6 mol\Rightarrow V=8,225 ml\Rightarrow V_{dd}=2V=16,45 ml$