If $y(x)$ satisfies the differential equation $y^{\prime}-y \tan x=2 x \sec x$ and $y(0)=0$, then

$y\left(\frac{\pi}{4}\right)=\frac{\pi^2}{8 \sqrt{2}}, y^{\prime}\left(\frac{\pi}{3}\right)=\frac{4 \pi}{3}+\frac{2 \pi^2}{3 \sqrt{3}}$

We have,

$\frac{d y}{d x}-(\tan x) y=2 x \sec x$            ......(i)

It is a linear differential equation with

Integrating factor = $e^{-\int \tan x d x}=e^{\log \cos x}=\cos x$

Multiplying (i) by integrating factor = $\cos x$ and integrating, we get

$\Rightarrow y \cos x=x^2+C$              ........(ii)

Putting $x=0$ and $y=0$, we get

$0=0+C \Rightarrow C=0$

Putting $C=0$ in (ii), we get

$y =x^2 \sec x$              ........(iii)

$\Rightarrow y' =2 x \sec x+x^2 \sec x \tan x$              ........(iv)

Putting $x=\frac{\pi}{4}$ in (iii) and (iv), we get

$y\left(\frac{\pi}{4}\right)=\frac{\pi^2}{8 \sqrt{2}}, y'\left(\frac{\pi}{4}\right)=\frac{\pi}{\sqrt{2}}+\frac{\pi^2}{8 \sqrt{2}}$

Putting $x=\frac{\pi}{3}$ in (iii) and (iv), we get

$y\left(\frac{\pi}{3}\right)=\frac{2 \pi^2}{9}, y'\left(\frac{\pi}{3}\right)=\frac{4 \pi}{3}+\frac{2 \pi^2}{3 \sqrt{3}}$