Given that
$$B = \begin{pmatrix}2&4\\3&2\end{pmatrix} + 13A_1$$
where $A_1\in M_2(\mathbb{Z}_2)$. We want to find all the possible choices of $B$ such that $\gcd{(\det{B},26)}=1$ and $\gcd{(\det{B},2)} = 1$.
So there are $2^4=16$ different choices of $B$ initially. Now how do we remove some of these choices. Is there any way to do this without considering them case by case?
In other words you want $\det(B)$ to be odd and not divisible by $13$. The "not divisible by $13$" isn't a problem, because $\det(B) \equiv \det\pmatrix{2 & 4\cr 3 & 2} \equiv -8 \mod 13$. If $A_1 = \pmatrix{a & b\cr c & d\cr}$, then $B \equiv \pmatrix{a & b\cr 1+c & d\cr} \mod 2$, so $\det(B) \equiv ad + b(1+c)\mod 2$. You want exactly one of $ad$ and $b(1+c)$ to be odd.