Ta có \(\left\{\begin{array}{l}\log _{8} a+\log _{4} b^{2}=5 \\ \log _{4} a^{2}+\log _{8} b=7\end{array} \Leftrightarrow\left\{\begin{array}{l}\frac{1}{3} \log _{2} a+\log _{2} b=5 \\ \log _{2} a+\frac{1}{3} \log _{2} b=7\end{array} \Leftrightarrow\left\{\begin{array}{l}\log _{2} a=6 \\ \log _{2} b=3\end{array} \Leftrightarrow\left\{\begin{array}{l}a=2^{6} \\ b=2^{3}\end{array}\right.\right.\right.\right.\).

Vậy \(\frac{a}{b}=\frac{2^{6}}{2^{3}}=2^{3}=8\).