5. If the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=\frac{2}{3}, a_{n+1}-a_{n}=\sqrt{\frac{2}{3}\left(a_{n+1}+a_{n}\right)}$, then $a_{2015}=$ $\qquad$ .
5. 1354080 .

From
$$
a_{n+1}-a_{n}=\sqrt{\frac{2}{3}\left(a_{n+1}+a_{n}\right)}
$$

squaring both sides, we get
$$
3\left(a_{n+1}-a_{n}\right)^{2}=2\left(a_{n+1}+a_{n}\right),
$$

and also
$$
3\left(a_{n}-a_{n-1}\right)^{2}=2\left(a_{n}+a_{n-1}\right),
$$

subtracting the two equations, we obtain
$$
3\left(a_{n+1}-a_{n-1}\right)\left(a_{n+1}-2 a_{n}+a_{n-1}\right)=2\left(a_{n+1}-a_{n-1}\right) .
$$

Given $a_{1}=\frac{2}{3}$ and the recurrence relation, we can find $a_{2}=2$. It is also easy to see from the recurrence relation that the sequence $\left\{a_{n}\right\}$ is a monotonically increasing sequence, so $a_{n+1}-a_{n-1} \neq 0$, hence $3\left(a_{n+1}-2 a_{n}+a_{n-1}\right)=2$, which means
$$
a_{n+1}-2 a_{n}+a_{n-1}=\frac{2}{3},
$$

or equivalently
$$
\left(a_{n+1}-a_{n}\right)-\left(a_{n}-a_{n-1}\right)=\frac{2}{3},
$$

thus 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 $\frac{2}{3}$, so
$$
a_{n+1}-a_{n}=\frac{4}{3}+\frac{2}{3}(n-1)=\frac{2}{3}(n+1) \text {, }
$$

therefore
$$
a_{n}=a_{1}+\frac{2}{3}(2+3+\cdots+n)=\frac{1}{3} n(n+1),
$$

so, $a_{2015}=\frac{1}{3} \times 2015 \times(2015+1)=1354080$.