Solution:

Scale so that $A_1A_2 = 2.$ A $90^\circ$ rotation about the center carries the whole configuration to itself, so $B_1B_3B_5B_7$ is a square and the distances $a = M_iB_i$ and $b = M_iB_{i-2}$ do not depend on $i.$ Both $B_3$ and $B_1$ lie on ray $R_3$ (at distances $a$ and $b$ from $M_3$), so $b - a = B_1B_3 = 2.$ Also, $R_1 \perp R_3$ makes triangle $M_1B_1M_3$ right-angled at $B_1,$ so $a^2 + b^2 = M_1M_3^2.$

Lines $A_1A_2$ and $A_3A_4$ meet at a point $C$ at a right angle, and triangle $A_2CA_3$ is an isosceles right triangle with legs $A_2C = A_3C = \sqrt{2},$ so $M_1C = M_3C = 1 + \sqrt{2}$ and $a^2 + b^2 = M_1M_3^2 = 2(1 + \sqrt{2})^2.$ Then $$(a + b)^2 = 2(a^2 + b^2) - (b - a)^2 = 4(1 + \sqrt{2})^2 - 4 = 8 + 8\sqrt{2}.$$

Since triangle $M_1CM_3$ is an isosceles right triangle and $A_3$ lies on segment $\overline{M_3C},$ we have $\angle A_3M_3M_1 = 45^\circ,$ while $\tan \angle M_1M_3B_1 = \frac{a}{b}$ from the right triangle. The tangent addition formula gives $$\tan \angle A_3M_3B_1 = \frac{1 + \frac{a}{b}}{1 - \frac{a}{b}} = \frac{a + b}{b - a}, \qquad \text{so} \quad \tan^2 \angle A_3M_3B_1 = \frac{8 + 8\sqrt{2}}{4} = 2 + 2\sqrt{2}.$$ Therefore $$\cos 2\angle A_3M_3B_1 = \frac{1 - \tan^2 \angle A_3M_3B_1}{1 + \tan^2 \angle A_3M_3B_1} = \frac{-1 - 2\sqrt{2}}{3 + 2\sqrt{2}} = -(1 + 2\sqrt{2})(3 - 2\sqrt{2}) = 5 - 4\sqrt{2} = 5 - \sqrt{32},$$ and $m + n = 5 + 32 = 37.$