8. Let the integer sequence $a_{1}, a_{2}, \cdots, a_{10}$ satisfy:
$$
a_{10}=3 a_{1}, a_{2}+a_{8}=2 a_{5} \text {, }
$$

and $a_{i+1} \in\left\{1+a_{i}, 2+a_{i}\right\}(i=1,2, \cdots, 9)$. Then the number of such sequences is $\qquad$
8. 80 .
$$
\begin{array}{l}
\text { Let } b_{i}=a_{i+1}-a_{i} \in\{1,2\}(i=1,2, \cdots, 9) \text {. } \\
\text { Then } 2 a_{1}=a_{10}-a_{1}=b_{1}+b_{2}+\cdots+b_{9}, \\
b_{2}+b_{3}+b_{4}=a_{5}-a_{2}=a_{8}-a_{5} \\
=b_{5}+b_{6}+b_{7} .
\end{array}
$$

Let $t$ represent the number of terms with value 2 among $b_{2}, b_{3}, b_{4}$. From equation (2), we know that $t$ is also the number of terms with value 2 among $b_{5}, b_{6}, b_{7}$, where $t \in\{0,1,2,3\}$.
Thus, the number of ways to choose $b_{2}, b_{3}, \cdots, b_{7}$ is
$$
\left(\mathrm{C}_{3}^{0}\right)^{2}+\left(\mathrm{C}_{3}^{1}\right)^{2}+\left(\mathrm{C}_{3}^{2}\right)^{2}+\left(\mathrm{C}_{3}^{3}\right)^{2}=20 \text {. }
$$

After fixing $b_{2}, b_{3}, \cdots, b_{7}$, we can arbitrarily assign values to $b_{8}, b_{9}$, which gives $2^{2}=4$ ways.

From equation (1), we know that $b_{1} \in\{1,2\}$ should be chosen such that $b_{1}+b_{2}+\cdots+b_{9}$ is even, and the choice of $b_{1}$ is unique, which also determines the value of the integer $a_{1}$.

Therefore, the sequence $b_{1}, b_{2}, \cdots, b_{9}$ uniquely corresponds to a sequence $a_{1}, a_{2}, \cdots, a_{10}$ that satisfies the conditions.
In summary, the number of sequences that satisfy the conditions is
$$
20 \times 4=80 \text {. }
$$