16. Given that $\vec{a}$ and $\vec{b}$ are non-zero vectors, and $\vec{a}+3 \vec{b}$ is perpendicular to $7 \vec{a}-5 \vec{b}$, $\vec{a}-4 \vec{b}$ is perpendicular to $7 \vec{a}-2 \vec{b}$, find the angle between $\vec{a}$ and $\vec{b}$.
16. From the problem, we have $\left\{\begin{array}{l}(\vec{a}+3 \vec{b}) \cdot(7 \vec{a}-5 \vec{b})=0 \\ (\vec{a}-4 \vec{b}) \cdot(7 \vec{a}-2 \vec{b})=0,\end{array}\right.$ which simplifies to $\left\{\begin{array}{l}7 \vec{a}+16 \vec{a} \cdot \vec{b}-15 \vec{b}^{2}=0 \\ 7 \vec{a}^{2}-30 \vec{a} \cdot \vec{b}+8 \vec{b}^{2}=0 \text {. }\end{array}\right.$ Subtracting the two equations gives $\vec{b}^{2}=2 \vec{a} \cdot \vec{b}$, and substituting this back into the equations yields $\vec{a}^{2}=2 \vec{a} \cdot \vec{b}$. The angle between $\vec{a}$ and $\vec{b}$ is $60^{\circ}$.