2. Given that the first $n$ terms sum of the arithmetic sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ are $S_{n}$ and $T_{n}$ respectively, and $\frac{S_{n}}{T_{n}}=\frac{3 n+2}{2 n+1}$. Then $\frac{a_{7}}{b_{5}}=$ $\qquad$ .
2. $\frac{41}{19}$.

From $\frac{S_{n}}{T_{n}}=\frac{3 n+2}{2 n+1}$, we get $\frac{S_{n}}{T_{n}}=\frac{\left(3 n^{2}+2 n\right) k}{\left(2 n^{2}+n\right) k}$.
Let $S_{n}=\left(3 n^{2}+2 n\right) k, T_{n}=\left(2 n^{2}+n\right) k$.
Then $a_{7}=S_{7}-S_{6}=41 k, b_{5}=T_{5}-T_{4}=19 k$.
Therefore, $\frac{a_{7}}{b_{5}}=\frac{41 k}{19 k}=\frac{41}{19}$.