8. Given the sequence $\left\{a_{n}\right\}$, the odd terms form an arithmetic sequence with a common difference of $d_{1}$, and the even terms form an arithmetic sequence with a common difference of $d_{2}$. For any $n \in \mathbf{N}^{*}$, it is true that $a_{n}<a_{n+1}$. If $a_{1}=1, a_{2}=2$, and the sum of the first 10 terms of the sequence $\left\{a_{n}\right\}$ is $S_{10}=75$, then $a_{8}=$ $\qquad$
$$
\begin{array}{l}
a_{1}+a_{3}+a_{5}+a_{7}+a_{9}=5 a_{1}+10 d_{1}=5+10 d_{1}, a_{2}+a_{4}+a_{6}+a_{8}+a_{10}=5 a_{2}+10 d_{2} \\
=10+10 d_{2} \Rightarrow 15+10\left(d_{1}+d_{2}\right)=75 \Rightarrow d_{1}+d_{2}=6 .
\end{array}
$$

For $n=2 k-1$, $a_{1}+(k-1) d_{1}1-d_{1}\end{array}\right.$ holds for any $k \in \mathbf{N}^{*}$ $\Rightarrow d_{1}=d_{2}=3$.
Therefore, $a_{8}=a_{2}+3 d_{2}=11$.