Let $y=y(x)$ be the solution of the differential equation $(\tan x)^{1 / 2} \mathrm{~d} y=\left(\sec ^3 x-(\tan x)^{3 / 2} y\right) \mathrm{d} x, 0 < x <\frac{\pi}{2}, y\left(\frac{\pi}{4}\right)=\frac{6 \sqrt{2}}{5}$. If $y\left(\frac{\pi}{3}\right)=\frac{4}{5} \alpha$, then $\alpha^4$ equals

$\_\_\_\_$ .

Let $y=y(x)$ be the solution of the differential equation $x \sin \left(\frac{y}{x}\right) d y=\left(y \sin \left(\frac{y}{x}\right)-x\right) d x, y(1)=\frac{\pi}{2}$ and let $\alpha=\cos \left(\frac{y\left(e^{12}\right)}{e^{12}}\right)$. Then the number of integral value of $p$, for which the equation $x^2+y^2-2 p x+2 p y+\alpha+2=0$ represents a circle of radius $r \leq 6$, is $\_\_\_\_$ .

Let $f$ be a twice differentiable function such that

$$ f(x)=\int_0^x \tan (t-x) d t-\int_0^x f(t) \tan t d t, x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) $$

Then $f^{\prime \prime}\left(\frac{\pi}{6}\right)+12 f^{\prime}\left(-\frac{\pi}{6}\right)+f\left(\frac{\pi}{6}\right)$ is equal to _______.

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 $\_\_\_\_$