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

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
$$
a_{7}=S_{7}-S_{6}=41 k, \quad b_{5}=T_{5}-T_{4}=19 k,
$$

Thus $\frac{a_{7}}{b_{5}}=\frac{41 k}{19 k}=\frac{41}{19}$.