8. The sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=\frac{2}{3}, a_{n+1}-a_{n}=\sqrt{\frac{2}{3}\left(a_{n+1}+a_{n}\right)} \text {. }
$$

Then $a_{2007}=$
8.1343352 .

Given $a_{n+1}-a_{n}=\sqrt{\frac{2}{3}\left(a_{n+1}+a_{n}\right)}$, squaring both sides yields $3\left(a_{n+1}-a_{n}\right)^{2}=2\left(a_{n+1}+a_{n}\right)$.

Also, $3\left(a_{n}-a_{n-1}\right)^{2}=2\left(a_{n}+a_{n-1}\right)$, subtracting the two equations gives
$$
\begin{array}{l}
3\left(a_{n+1}-a_{n-1}\right)\left(a_{n+1}-2 a_{n}+a_{n-1}\right) \\
=2\left(a_{n+1}-a_{n-1}\right) .
\end{array}
$$

Given $a_{1}=\frac{2}{3}, a_{n+1}-a_{n}=\sqrt{\frac{2}{3}\left(a_{n+1}+a_{n}\right)}$, we find $a_{2}=2$.

From the recursive relation, it is easy to see that the sequence $\left\{a_{n}\right\}$ is a monotonically increasing sequence, so $a_{n+1}-a_{n-1} \neq 0$.
Hence $3\left(a_{n+1}-2 a_{n}+a_{n-1}\right)=2$
$$
\Rightarrow\left(a_{n+1}-a_{n}\right)-\left(a_{n}-a_{n-1}\right)=\frac{2}{3} \text {. }
$$

Therefore, the sequence $\left\{a_{n+1}-a_{n}\right\}$ is an arithmetic sequence with the first term $a_{2}-a_{1}=\frac{4}{3}$ and common difference $\frac{2}{3}$.
Thus, $a_{n+1}-a_{n}=\frac{4}{3}+\frac{2}{3}(n-1)=\frac{2}{3}(n+1)$.
Hence, $a_{n}=a_{1}+\frac{2}{3}(2+3+\cdots+n)=\frac{1}{3} n(n+1)$.
Therefore, $a_{2007}=1343352$.