1. Given differential equation

$2\cos x\,\frac{dy}{dx}=\sin 2x-4y\sin x, \qquad x\in\left(0,\frac{\pi}{2}\right)$

We use $\sin 2x=2\sin x\cos x$:

$2\cos x\,y'=2\sin x\cos x-4y\sin x$

Divide by $2\cos x$ (valid since $x\in(0,\pi/2)$, so $\cos x>0$):

$y'=\sin x-2y\tan x$

So,

$y'+2y\tan x=\sin x$

2. Solve the linear differential equation

This is of the form

$y'+P(x)y=Q(x), \qquad P(x)=2\tan x$

Integrating factor:

$\text{I.F.}=e^{\int 2\tan x\,dx}=e^{2\int \tan x\,dx}$

Since $\int \tan x\,dx=\ln(\sec x)$,

$\text{I.F.}=e^{2\ln(\sec x)}=\sec^2 x$

Multiply the equation by $\sec^2 x$:

$\sec^2 x\,y'+2\tan x\sec^2 x\,y=\sin x\sec^2 x$

Left side is:

$\frac{d}{dx}(y\sec^2 x)=\sin x\sec^2 x$

Now,

$\sin x\sec^2 x=\frac{\sin x}{\cos^2 x}=\tan x\sec x$

Hence,

$y\sec^2 x=\int \tan x\sec x\,dx=\sec x+C$

Therefore,

$y=(\sec x+C)\cos^2 x=\cos x+C\cos^2 x$

3. Use the condition $y\left(\frac{\pi}{3}\right)=0$

At $x=\frac{\pi}{3}$,

$0=y\left(\frac{\pi}{3}\right)=\cos\frac{\pi}{3}+C\cos^2\frac{\pi}{3}$

$0=\frac12+C\cdot\frac14$

$\frac{C}{4}=-\frac12 \implies C=-2$

So the solution is

$y=\cos x-2\cos^2 x$

4. Find $y\left(\frac{\pi}{4}\right)$

Since $\cos\frac{\pi}{4}=\frac{\sqrt2}{2}$ and $\cos^2\frac{\pi}{4}=\frac12$,

$y\left(\frac{\pi}{4}\right)=\frac{\sqrt2}{2}-2\cdot\frac12=\frac{\sqrt2}{2}-1$

5. Find $y'\left(\frac{\pi}{4}\right)$

Differentiate:

$y=\cos x-2\cos^2 x$

$y'=-\sin x-2\cdot 2\cos x(-\sin x)=-\sin x+4\sin x\cos x$

At $x=\frac{\pi}{4}$,

$\sin\frac{\pi}{4}=\cos\frac{\pi}{4}=\frac{\sqrt2}{2}, \qquad 4\sin\frac{\pi}{4}\cos\frac{\pi}{4}=4\cdot\frac12=2$

Thus,

$y'\left(\frac{\pi}{4}\right)=-\frac{\sqrt2}{2}+2$

6. Compute the required value

$y'\left(\frac{\pi}{4}\right)+y\left(\frac{\pi}{4}\right)=\left(2-\frac{\sqrt2}{2}\right)+\left(\frac{\sqrt2}{2}-1\right)=1$

So the required integer is

$\boxed{1}$

7. Comparison with stored answer

Stored correct answer: $1$

Our derived answer is also $1$, so they agree.