7. Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{0}=0$,
$$
a_{n+1}=\frac{8}{5} a_{n}+\frac{6}{5} \sqrt{4^{n}-a_{n}^{2}}(n \geqslant 0, n \in \mathbf{N}) \text {. }
$$

Then $a_{10}=$ . $\qquad$
7. $\frac{24576}{25}$.

Notice, $a_{n+1}=2\left(\frac{4}{5} a_{n}+\frac{3}{5} \sqrt{4^{n}-a_{n}^{2}}\right)$.
Let $a_{n}=2^{n} \sin \alpha_{n}$.
Let $\theta$ satisfy $\sin \theta=\frac{3}{5}, \cos \theta=\frac{4}{5}$.
Then $\sin \alpha_{n+1}=\sin \alpha_{n} \cdot \cos \theta+\sin \theta \cdot \mid \cos \alpha_{n}$ |
$$
=\left\{\begin{array}{l}
\sin \left(\alpha_{n}+\theta\right), \cos \alpha_{n}>0 ; \\
\sin \left(\alpha_{n}-\theta\right), \cos \alpha_{n}<0 .
\end{array}\right.
$$

Also, $\cos 45^{\circ}<\frac{4}{5}<\cos 30^{\circ}$
$$
\Rightarrow 30^{\circ}<\theta<45^{\circ} \text {. }
$$

Thus, $a_{n}=\left\{\begin{array}{l}2^{n} \sin n \theta, n=0,1,2 ; \\ 2^{n} \sin 2 \theta, n \text { is greater than } 2 \text { and even; } \\ 2^{n} \sin 3 \theta, n \text { is greater than } 1 \text { and odd. }\end{array}\right.$ Therefore, $a_{10}=2^{10} \sin 2 \theta$
$$
=1024 \times \frac{24}{25}=\frac{24576}{25} .
$$