I'll just do the last two integrals you got to:
For the first you can apply IBP choosing $u=x$ and $\mathrm dv=\dfrac{\mathrm dx}{1+\cos x}$:
$$\begin{aligned}
I_1\equiv\int_0^{\pi/2}\!\!\!\!\frac{x}{1+\cos x}\,\mathrm dx&\overset{(\ast)^1}{=}\underbrace{\left[x\tan\left(\frac{x}{2}\right)\right|_0^{\pi/2}}_{\pi/2}-\int_0^{\pi/2}\!\!\!\!\tan\left(\frac{x}{2}\right)\,\mathrm dx\\
&\overset{(\ast)^2}{=}\frac{\pi}{2}-\left[-2\ln\left(\cos\left(\frac{x}{2}\right)\right)\right|_0^{\pi/2}=\frac{\pi}{2}-\ln 2
\end{aligned}$$
As for the other one you can apply IBP yet again by choosing $u=x$ and $\mathrm dv=\dfrac{\sin x\ \mathrm dx}{1+\cos x}$:
$$\begin{aligned}
I_2\equiv\int_0^{\pi/2}\!\!\!\!\frac{x\sin x}{1+\cos x}\,\mathrm dx&\overset{(\ast)^3}{=}\underbrace{\left[-x\ln(1+\cos x)\right|_0^{\pi/2}}_0+\int_0^{\pi/2}\!\!\!\!\ln(1+\cos x)\,\mathrm dx\\
&=2\mathrm G-\frac{\pi}{2}\ln2\\
\end{aligned}$$
where $\mathrm G$ is Catalan's constant and to compute the later integral you can adress to this MSE post.
Thus,
$$
\int_0^{\pi/4}\!\!\!\!\frac{x\cos^2(2x)}{(1+\sin(2x))(1+\cos(2x))}\,\mathrm dx=\frac{1}{4}(I_2-I_1)=\boxed{\frac{\mathrm G}{2}+\frac{2-\pi}{8}\ln2-\frac{\pi}{8}}
$$
$(\ast)^1:$
$$\int\! \dfrac{\mathrm dx}{1+\cos x}=\int\dfrac{1}{1+\dfrac{1-t^2}{1+t^2}}\frac{2\ \mathrm dt}{1+t^2}=\int\mathrm dt=t=\tan\left(\frac{x}{2}\right)$$
where here we used Weierstrass substitution $\left(t=\tan\left(\frac{x}{2}\right)\right)$ and so $\cos x=\frac{1-t^2}{1+t^2}$ and $\mathrm dx=\frac{2\mathrm dt}{1+t^2}$.
$(\ast)^2:$
$$\int \tan\left(\frac{x}{2}\right)\,\mathrm dx=\int\frac{\sin\left(\frac{x}{2}\right)}{\cos\left(\frac{x}{2}\right)}\,\mathrm dx=-2\int \frac{\mathrm du}{u}=-2\ln u=-2\ln\left(\cos\left(\frac{x}{2}\right)\right)$$
where we did $u=\cos\left(\frac{x}{2}\right)\iff -2\mathrm du=\sin\left(\frac{x}{2}\right)\,\mathrm dx$.
$(\ast)^3:$
$$\int\! \dfrac{\sin x\ \mathrm dx}{1+\cos x}=-\int\frac{\mathrm du}{u}=-\ln(1+\cos x)$$
where we did $u=1+\cos x\iff \mathrm du=-\sin x\,\mathrm dx$.