jmc

Define a new sequence $(b_n)$ such that $a_n=2^n{b}_n$ for each $n.$ Then the recurrence becomes $$2^{n+1}{b}_{n+1}=\frac{8}{5}\cdot 2^n{b}_n+\frac{6}{5}\sqrt{4^n-4^n{b}_n^2}=\frac{8}{5}\cdot 2^n{b}_n+\frac{6}{5}\cdot 2^n\sqrt{1-b_n^2},$$or, dividing by $2^{n+1},$ $$b_{n+1}=\frac{4}{5}{b}_n+\frac{3}{5}\sqrt{1-b_n^2}.$$Compute by hand: $$\begin{array}{rl}{b}_1& =\frac{3}{5}\\ b_2& =\frac{4}{5}\cdot \frac{3}{5}+\frac{3}{5}\sqrt{1-{\left(\frac{3}{5}\right)}^2}=\frac{24}{25}\\ b_3& =\frac{4}{5}\cdot \frac{24}{25}+\frac{3}{5}\sqrt{1-{\left(\frac{24}{25}\right)}^2}=\frac{96}{125}+\frac{3}{5}\cdot \frac{7}{25}=\frac{117}{125}\\ b_4& =\frac{4}{5}\cdot \frac{117}{125}+\frac{3}{5}\sqrt{1-{\left(\frac{117}{125}\right)}^2}=\frac{468}{625}+\frac{3}{5}\cdot \frac{44}{125}=\frac{600}{625}=\frac{24}{25}\end{array}$$Since $b_2=b_4,$ the sequence $(b_n)$ starts to repeat with period $2.$ Thus, $b_{10}=b_2=\frac{24}{25},$ so $a_{10}=2^{10}{b}_{10}=\frac{2^{10}\cdot 24}{25}=\boxed{\frac{24576}{25}}.$