13. (15 points) In the sequence $\left\{a_{n}\right\}$,
$$
a_{n}=2^{n} a+b n-80\left(a 、 b \in \mathbf{Z}_{+}\right) \text {. }
$$

It is known that the minimum value of the sum of the first $n$ terms $S_{n}$ is obtained only when $n=6$, and $7 \mid a_{36}$. Find the value of $\sum_{i=1}^{12}\left|a_{i}\right|$.
Three, 13. Notice that, $\left\{a_{n}\right\}$ is an increasing sequence.
From the given, $a_{6}0$, that is,
$$
64 a+6 b-800 \text {. }
$$

Combining $a, b \in \mathbf{Z}_{+}$, we get
$$
a=1, b=1 \text { or } 2 \text {. }
$$

Also, $a_{36}=2^{36}+36 b-80$
$$
\equiv 1+b-3 \equiv 0(\bmod 7) .
$$

Thus, $b=2$.
Therefore, $a_{n}=2^{n}+2 n-80$.
Hence, $\left|a_{1}\right|+\left|a_{2}\right|+\cdots+\left|a_{12}\right|$
$$
\begin{array}{l}
=-\left(a_{1}+a_{2}+\cdots+a_{6}\right)+\left(a_{7}+a_{8}+\cdots+a_{12}\right) \\
=S_{12}-2 S_{6}=8010 .
\end{array}
$$