1. Let $\overrightarrow a \cdot \overrightarrow a = |\overrightarrow a|^2 = A, \qquad \overrightarrow b \cdot \overrightarrow b = |\overrightarrow b|^2 = B, \qquad \overrightarrow a \cdot \overrightarrow b = C.$

2. Use the condition that perpendicular vectors have zero dot product.

   From $\left(\overrightarrow a+3\overrightarrow b\right) \perp \left(7\overrightarrow a-5\overrightarrow b\right),$ we get $\left(\overrightarrow a+3\overrightarrow b\right)\cdot\left(7\overrightarrow a-5\overrightarrow b\right)=0.$

   Expanding: $7A-5C+21C-15B=0$ $7A+16C-15B=0. \qquad (1)$

3. From $\left(\overrightarrow a-4\overrightarrow b\right) \perp \left(7\overrightarrow a-2\overrightarrow b\right),$ we get $\left(\overrightarrow a-4\overrightarrow b\right)\cdot\left(7\overrightarrow a-2\overrightarrow b\right)=0.$

   Expanding: $7A-2C-28C+8B=0$ $7A-30C+8B=0. \qquad (2)$

4. Subtract (2) from (1): $\left(7A+16C-15B\right)-\left(7A-30C+8B\right)=0$ $46C-23B=0$ $2C-B=0$ $B=2C. \qquad (3)$

5. Put (3) into (2): $7A-30C+8(2C)=0$ $7A-14C=0$ $A=2C. \qquad (4)$

6. Therefore, $A=B=2C.$

   Now, \cos\theta=\frac{\overrightarrow a\cdot\overrightarrow b}{|\overrightarrow a||\overrightarrow b|}=rac{C}{\sqrt{AB}}=rac{C}{\sqrt{(2C)(2C)}}=\frac{C}{2C}=\frac12.

7. Hence, $\theta=60^\circ.$

Therefore, the required angle between $\overrightarrow a$ and $\overrightarrow b$ is $60^\circ$.