According to this post
$$Q=\int_0^1\frac{\ln(x)}{1-x/2}\int_0^{x/2}\frac{\ln(1-t)}{t}\,dt\,dx=\frac{1}{8}\zeta \left( 4 \right) - \frac{1}{2}\zeta \left( 2 \right){\ln ^2}(2) + \frac{1}{{12}}{\ln ^4}(2) + \frac{1}{2}\zeta \left( 3 \right)\ln (2)$$
however I had a hard time proving it several years ago when I first looked at it. I had reduced it to
$$Q=I+J-\ln(2)K+\frac{\pi^4}{36}-\frac{\pi^2}{6}\ln^2(2)+\frac{\pi^4}{36}\ln^2(2)+\frac{\ln^4(2)}{3}-\frac{7}{3}\ln(2)\zeta(3)$$
where
$$I=\int_0^1\frac{\ln(x)}{1-x/2}\text{Li}_2\left(1-\frac{x}{2}\right)dx$$
$$J=\int_0^1\frac{\ln^2(x)\ln(2-x)}{1-x/2}dx$$
$$K=\int_0^1\frac{\ln(x)\ln(2-x)}{1-x/2}dx.$$
But then I got stuck.
For reference on how I proceeded I tried using the polylogarithmic identity
$$\text{Li}_2(1-s)=-\text{Li}_2(s)-\ln(1-s)\ln(s)+\frac{\pi^2}{6}$$
and used some results form the similar integral
$$\begin{align*} \int_0^1\frac{\ln(x)}{1-\frac{x}{2}}\left (\int_1^{\frac{x}{2}}\frac{1}{t}dt \right )dx&=\int_0^1\frac{\ln(x)\ln(x/2)}{1-\frac{x}{2}}dx \\ &=\int_0^1\frac{\ln^2(x)}{1-\frac{x}{2}}dx-\ln(2)\int_0^1\frac{\ln(x)}{1-\frac{x}{2}}dx \\ &=\sum_{n=0}^\infty\int_0^1\left ( \frac{x}{2} \right )^n\ln^2(x)dx-\ln(2)\sum_{n=0}^\infty\int_0^1\left ( \frac{x}{2} \right )^n\ln(x)dx \\ &=\sum_{n=0}^\infty\frac{2}{2^n(n+1)^3}+\ln(2)\sum_{n=0}^\infty\frac{1}{2^n(n+1)^2} \\ &=4\text{Li}_3\left ( \frac{1}{2} \right )+2\ln(2)\text{Li}_2\left(\frac{1}{2} \right ) \\ &= \frac{7}{3}\zeta(3)-\frac{1}{3}\ln^3(2)-\frac{\pi^2}{6}\ln(2). \end{align*}$$
To reiterate my question I'm looking for a technique to evaluate Q. The main issue is finding methods to evaluate $I$ and $K$. I think if we had a method to evaluate $K$ it could be modified to get $J$s value as well. If anyone has an alternate approach to get the value of Q then that'd be cool too.
Update: We can evaluate $I$ pretty easily,
$$\begin{align*} \int_0^1\frac{\ln(x)}{1-x/2}\text{Li}_2\left(1-\frac{x}{2}\right)dx&=-2\int_{1}^{1/2}\frac{\ln(2-2u)}{u}\text{Li}_2(u)du,\;\text{let }u=1-\frac{x}{2}\\ &=-2\left(\int_1^{1/2}\frac{\ln(2)}{u}\text{Li}_2(u)du+\int_1^{1/2}\frac{\ln(1-u)}{u}\text{Li}_2(u)du\right)\\ &=-2\ln(2)\left[\text{Li}_3(u)\right]\rvert_{1}^{1/2}-2\left[-\text{Li}_2^2(u)/2\right]\rvert_{1}^{1/2}\\ &=\frac{1}{4}\zeta(3)\ln(2)+\frac{\pi^2}{12}\ln^2(2)-\frac{\ln^4(2)}{12}-\frac{\pi^4}{48}. \end{align*}$$
And $K$ is a boring exercise in integrating by parts,
$$\begin{align*} \int_0^1\frac{\ln(x)\ln(2-x)}{1-x/2}dx&=\left[-2\text{Li}_3(1-x/2)+2\text{Li}_2(1-x/2)\ln(2-x)-\ln(2)\ln^2(2-x)\right]\rvert_0^1\\ &=\frac{1}{4}\zeta(3)+\frac{2}{3}\ln^3(2)-\frac{\pi^2}{6}\ln(2). \end{align*}$$
I will use my own notation for your $Q$ integral. \begin{align} \mathfrak{I} &= \int\limits_{0}^{1} \dfrac{\ln x}{1-x/2}\left(\int\limits_{0}^{x/2} \dfrac{\ln \left(1-t\right)}{t}\,\mathrm{d}t\right)\,\mathrm{d}x \\ &= -\int\limits_{0}^{1} \dfrac{\ln x}{1-x/2}\operatorname{Li}_2\left(\dfrac{x}{2}\right)\,\mathrm{d}x \\ &= -2\int\limits_{0}^{1/2} \dfrac{\ln (2x)}{1-x}\operatorname{Li}_2(x)\,\mathrm{d}x \\ &= -2\ln 2\underbrace{\int\limits_{0}^{1/2} \dfrac{\operatorname{Li}_2(x)}{1-x}\,\mathrm{d}}_{\mathfrak{I}_1}-2\underbrace{\int\limits_{0}^{1/2} \dfrac{\ln x}{1-x}\operatorname{Li}_2(x)\,\mathrm{d}x}_{\mathfrak{I}_2} \end{align} $\mathfrak{I}_1$ can be evaluated as follows: \begin{align} \mathfrak{I}_1 &= -\ln\left(1-x\right)\operatorname{Li}_2(x)\Bigg\vert_0^{1/2}-\int\limits_{0}^{1/2} \dfrac{\ln^2 \left(1-x\right)}{x}\,\mathrm{d}x \\ &= \ln 2\operatorname{Li}_2\left(\dfrac{1}{2}\right)-\int\limits_{1/2}^{1} \dfrac{\ln^2 x}{1-x}\,\mathrm{d}x \\ &= \ln 2\operatorname{Li}_2\left(\dfrac{1}{2}\right)-\sum\limits_{n=0}^{\infty}\,\int\limits_{1/2}^{1} x^n\ln^2 x\,\mathrm{d}x \\ &= \ln 2\operatorname{Li}_2\left(\dfrac{1}{2}\right)-\sum\limits_{n=0}^{\infty}\left.\left(\dfrac{x^{n+1}}{n+1}\ln^2 x-2\dfrac{x^{n+1}}{\left(n+1\right)^2}\ln x+2\dfrac{x^{n+1}}{\left(n+1\right)^3}\right)\right\vert_{1/2}^{1} \\ &= \ln 2\operatorname{Li}_2\left(\dfrac{1}{2}\right)-2\operatorname{Li}_2(1)+\ln^3 2 + 2\ln 2\operatorname{Li}_2\left(\dfrac{1}{2}\right)+2\operatorname{Li}_3\left(\dfrac{1}{2}\right) \\ &= -\dfrac{1}{4}\zeta(3)-\dfrac{1}{6}\ln^3 2 + \dfrac{1}{12}\pi^2\ln 2 \end{align} So
For $\mathfrak{I}_2$ we note that: \begin{align} \mathfrak{I}_2 &= \operatorname{Li}_2\left(1-x\right)\operatorname{Li}_2(x)\Bigg\vert_0^{1/2}+\int\limits_{0}^{1/2} \dfrac{\ln \left(1-x\right)}{x}\operatorname{Li}_2\left(1-x\right)\,\mathrm{d}x \\ &= \operatorname{Li}_2^2\left(\dfrac{1}{2}\right)+\int\limits_{1/2}^{1} \dfrac{\ln x}{1-x}\operatorname{Li}_2(x)\,\mathrm{d}x \\ &= \operatorname{Li}_2^2\left(\dfrac{1}{2}\right)+\int\limits_{0}^{1} \dfrac{\ln x}{1-x}\operatorname{Li}_2(x)\,\mathrm{d}x - \mathfrak{I}_2 \end{align} But last integral: \begin{align} \int\limits_{0}^{1} \dfrac{\ln x}{1-x}\operatorname{Li}_2(x)\,\mathrm{d}x &= -\ln \left(1-x\right)\ln x\operatorname{Li}_2(x)\Bigg\vert_{0}^{1}+\int\limits_{0}^{1} \dfrac{\ln \left(1-x\right)}{x}\operatorname{Li}_2(x)\,\mathrm{d}x-\int\limits_{0}^{1} \dfrac{\ln x\ln^2 \left(1-x\right)}{x}\,\mathrm{d}x \\ &= -\dfrac{1}{2}\operatorname{Li}_2^2(x)\Bigg\vert_{0}^{1}-\dfrac{\partial^3}{\partial x \partial y^2}\operatorname{B}(x,y)\Bigg\vert_{(0^+,1)} \\ &= -\dfrac{1}{120}\pi^4 \end{align} , where $\operatorname{B}(\cdot, \cdot)$ represent beta function. Whereupon $\mathfrak{I}_2$ can be easily calculated
Connecting parts together, we have: \begin{align} \mathfrak{I} &= -2\mathfrak{I}_1 \ln 2 - 2\mathfrak{I}_2 \\ &= \left(\dfrac{1}{2}\zeta(3)\ln 2 + \dfrac{1}{3}\ln^4 2 - \dfrac{1}{6}\pi^2\ln^2 2\right) + \left(\dfrac{1}{720}\pi^4 - \dfrac{1}{4}\ln^4 2 + \dfrac{1}{12}\pi^2\ln^2 2\right) \\ &= \dfrac{1}{720}\pi^4 + \dfrac{1}{2}\zeta(3)\ln 2 - \dfrac{1}{12}\pi^2\ln^2 2 + \dfrac{1}{12}\ln^4 2 \end{align}