jmc

We can use the Euclidean Algorithm. $$\begin{array}{rl}& \text{gcd}(2a^2+29a+65,a+13)\\ & =\text{gcd}(2a^2+29a+65-(a+13)(2a+3),a+13)\\ & =\text{gcd}(2a^2+29a+65-(2a^2+29a+39),a+13)\\ & =\text{gcd}(26,a+13).\end{array}$$Since $a$ is an odd multiple of $1183$, which is an odd multiple of $13$, $a+13$ must be an even multiple of $13$. This means that $26$ is a divisor of $a+13$, so the greatest common divisor is $\boxed{26}$.