If the solution curve $y=f(x)$ of the differential equation

$$ \left(x^2-4\right) y^{\prime}-2 x y+2 x\left(4-x^2\right)^2=0, x>2, $$

passes through the point $(3,15)$, then the local maximum value of $f$ is $\_\_\_\_$

Let $f$ be a twice differentiable non-negative function such that $(f(x))^2=25+\int_0^x\left((f(\mathrm{t}))^2+\left(f^{\prime}(\mathrm{t})\right)^2\right) \mathrm{dt}$. Then the mean of $f\left(\log _{\mathrm{e}}(1)\right), f\left(\log _{\mathrm{e}}(2)\right), \ldots . ., f\left(\log _{\mathrm{e}}(625)\right)$ is equal to $\_\_\_\_$ .

If $y=y(x)$ is the solution of the differential equation, $\sqrt{4-x^2} \frac{\mathrm{~d} y}{\mathrm{~d} x}=\left(\left(\sin ^{-1}\left(\frac{x}{2}\right)\right)^2-y\right) \sin ^{-1}\left(\frac{x}{2}\right),-2 \leq x \leq 2, y(2)=\frac{\pi^2-8}{4}$, then $y^2(0)$ is equal to ___________.