The even integers less than $12$ in the domain are $2,4,6,8,10$. There are $5$ such integers.
Exactly two of them must be assigned $1$, so this can be done in $\binom{5}{2}=10$ ways.
The odd integers greater than $3$ in the domain are $5,7,9,11$. There are $4$ such integers.
Exactly three of them must be assigned $0$, so this can be done in $\binom{4}{3}=4$ ways.
The remaining unrestricted elements are $1,3,12$. Each of these can be assigned either $0$ or $1$, so there are $2^3=8$ ways.
Hence the total number of functions is $\binom{5}{2}\binom{4}{3}2^3=10\cdot 4\cdot 8=320$.