Let $\alpha$ be the common central angle at $C$ in the second circle; then $\angle ACE = 5\alpha$ is simultaneously an inscribed angle of the first circle

Express every arc of both circles in terms of $\alpha;$ the condition $\angle ABD - \angle AHG = 12^\circ$ pins down $\alpha$

Let $\alpha = \angle ECF$ $= \angle FCG$ $= \angle GCH$ $= \angle HCI$ $= \angle ICA,$ the common central angle of the second circle, so $\angle ACE = 5\alpha.$ Since $C$ also lies on the first circle, $\angle ACE$ is an inscribed angle there, so the arc $AE$ not containing $C$ measures $10\alpha,$ and each of the four equal arcs $AB,$ $BC,$ $CD,$ $DE$ measures $\frac{360^\circ - 10\alpha}{4} = 90^\circ - \frac{5\alpha}{2}.$

Angle $ABD$ subtends the arc $AD$ not containing $B,$ which is $360^\circ - 3\left(90^\circ - \frac{5\alpha}{2}\right),$ so $\angle ABD = 45^\circ + \frac{15\alpha}{4}.$ Angle $AHG$ subtends the second circle’s arc $AG$ not containing $H,$ which is $360^\circ - 3\alpha,$ so $\angle AHG = 180^\circ - \frac{3\alpha}{2}.$ The given condition reads $$\begin{aligned} &\left(45^\circ + \frac{15\alpha}{4}\right) - \left(180^\circ - \frac{3\alpha}{2}\right) \\ &= \frac{21\alpha}{4} - 135^\circ = 12^\circ, \end{aligned}$$ so $\alpha = 28^\circ.$

Finally, $\angle BAE$ subtends the first circle’s arc $BCDE = 3\left(90^\circ - \frac{5\alpha}{2}\right) = 60^\circ,$ giving $\angle BAE = 30^\circ,$ and $\angle EAG$ subtends the second circle’s arc $EFG = 2\alpha,$ giving $\angle EAG = 28^\circ.$ Hence $\angle BAG = \angle BAE + \angle EAG$ $= 30^\circ + 28^\circ = 58^\circ.$