(2) Given that $S_{n}$ and $T_{n}$ are the sums of the first $n$ terms of the arithmetic sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$, respectively, and
$$
\frac{S_{n}}{T_{n}}=\frac{2 n+1}{4 n-2}(n=1,2, \cdots) \text {, then } \frac{a_{10}}{b_{3}+b_{18}}+\frac{a_{11}}{b_{6}+b_{15}}=
$$
$\qquad$
(2) $\frac{41}{78}$ Hint: Since $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ are arithmetic sequences, $\frac{S_{n}}{T_{n}}=\frac{2 n+1}{4 n-2}$, therefore
$$
\begin{aligned}
\frac{a_{10}}{b_{3}+b_{18}}+\frac{a_{11}}{b_{6}+b_{15}} & =\frac{a_{10}}{b_{1}+b_{20}}+\frac{a_{11}}{b_{1}+b_{20}}=\frac{a_{10}+a_{11}}{b_{1}+b_{20}} \\
& =\frac{a_{1}+a_{20}}{b_{1}+b_{20}}=\frac{S_{20}}{T_{20}}=\frac{41}{78} .
\end{aligned}
$$