$n_{H_2}= \frac{3,36}{22,4}= 0,15 mol$
$n_{C_2H_5OH}= a; n_{C_6H_5OH}= b$
$2C_2H_5OH+ 2Na \to 2C_2H_5ONa+ H_2$
$2C_6H_5OH+ 2Na \to 2C_6H_5ONa+ H_2$
=> $0,5a+0,5b=0,15$ (1)
$C_6H_5OH+ 3Br_2 \to C_6H_2Br_3OH \downarrow + 3HBr$
=> $331b= 19,86$ (2)
(1)(2) => $a=0,24; b=0,06$
$\%m_{C_2H_5OH}= \frac{0,24.46.100}{0,24.46+0,06.94}= 66,19\%$
$\%m_{C_6H_5OH}= 33,81\%$