Given the vector relation $(\mathbf{a}+3 \mathbf{b}) \perp(7 \mathbf{a}-5 \mathbf{b})$, $(\mathbf{a}-4 \mathbf{b}) \perp(7 \mathbf{a}-2 \mathbf{b})$, find the angle between $\mathbf{a}$ and $\mathbf{b}$.

$$\begin{array}{c}
\text{Solving process: } \qquad \because(\mathbf{a}+3 \mathbf{b}) \cdot(7 \mathbf{a}-5\mathbf{b}) = 0,(\mathbf{a}-4 \mathbf{b}) \cdot(7 \mathbf{a}-2 \mathbf{b}) = 0 \\
\therefore \quad 7 \mathbf{a}^{2}+16 \mathbf{a} \cdot \mathbf{b}-15 \mathbf{b}^{2} = 0,7 \mathbf{a}^{2}-30 \mathbf{a} \cdot \mathbf{b}+8 \mathbf{b}^{2} = 0\\
\quad \mathbf{a}^{2} = \mathbf{b}^{2} = 2 \mathbf{a} \cdot \mathbf{b} \\
\Rightarrow \cos \langle \mathbf{a}, \mathbf{b}\rangle = \frac{\mathbf{a} \cdot \mathbf{b}}{\mathbf{a}|\cdot \mathbf{b}|} = \frac{1}{2}
\end{array}$$