$C_6H_5OH+3Br_2\rightarrow Br_3C_6H_2OH+3HBr$$C_6H_5CH=CH_2+Br_2\rightarrow C_6H_5CHBr-CH_2Br$
$*)\begin{cases}C_6H_5OH:x mol \\ C_6H_5CH=CH_2:y mol \end{cases}\Rightarrow \begin{cases}3x+y=\frac{300.3,2}{160.100} \\ 4x=(n_{Br_3C_6H_2OH}+n_{HBr})=\frac{14,4.1,11.10}{40.100} \end{cases}$
$\Rightarrow \begin{cases}x\simeq 0,02 mol\\ y\simeq 0,04 mol\end{cases}\Rightarrow \begin{cases}m_{C_6H_5OH}=1,88 gam \\ m_{C_6H_5CH=CH_2}=4,16 gam \end{cases}\Rightarrow \begin{cases}\%C_6H_5OH=31,13\% \\ \%C_8H_8=68,87\% \end{cases}$