Point $A,B,C,D,$ and $E$ are equally spaced on a minor arc of a circle. Points $E,F,G,H,I$ and $A$ are equally spaced on a minor arc of a second circle with center $C$ as shown in the figure below. The angle $\angle ABD$ exceeds $\angle AHG$ by $12^\circ$. Find the degree measure of $\angle BAG$.
Let $O$ be the center of the circle with $ABCDE$ on it. 
Let $x$ be the degree measurement of $\overarc{ED}=\overarc{DC}=\overarc{CB}=\overarc{BA}$ in circle $O$
and $y$ be the degree measurement of $\overarc{EF}=\overarc{FG}=\overarc{GH}=\overarc{HI}=\overarc{IA}$ in circle $C$.
$\angle ECA$ is, therefore, $5y$ by way of circle $C$ and \[\frac{360-4x}{2}=180-2x\] by way of circle $O$.
$\angle ABD$ is $180 - \frac{3x}{2}$ by way of circle $O$, and \[\angle AHG = 180 - \frac{3y}{2}\] by way of circle $C$.
This means that:
\[180-\frac{3x}{2}=180-\frac{3y}{2}+12\]
which when simplified yields \[\frac{3x}{2}+12=\frac{3y}{2}\] or \[x+8=y\]
Since:
\[5y=180-2x\] and \[5x+40=180-2x\]
So:
\[7x=140\Longleftrightarrow x=20\]
\[y=28\]
$\angle BAG$ is equal to $\angle BAE$ + $\angle EAG$, which equates to $\frac{3x}{2} + y$.
Plugging in yields $30+28$, or $\boxed{058}$.