jmc

The only ways to write 13 as the product of two integers is as $13=1\times 13$ or $13=(-1)\times (-13)$. We take these two cases separately.

In the case $13=1\times 13$, we must have $a_4=1$ and $a_5=13$, since the sequence is increasing. Then the common difference is $13-1=12$, so $a_3=a_4-12=1-12=-11$, and $a_6=a_5+12=13+12=25$, so $a_3{a}_6=(-11)\cdot 25=-275$.

In the case $13=(-1)\times (-13)$, we must have $a_4=-13$ and $a_5=-1$. Then the common difference is $-1-(-13)=12$, so $a_3=a_4-12=-13-12=-25$, and $a_6=a_5+12=(-1)+12=11$, so $a_3{a}_6=(-25)\cdot 11=-275$.

Hence, $a_3{a}_6=\boxed{-275}$.