$1)$ Write $\text{Li}_2(1-\frac{1}{x})$ in terms of $\text{Li}_2(x)$ and logarithms by considering its integral representation and suitable changes of variables.
Attempt: The di-log is defined as $$\text{Li}_2(z) = \int_0^z \frac{dt}{t} \text{Li}_{1}(t) = -\int_0^z \frac{dt}{t} \text{log}(1-t) .$$ I'm just trying to see how to start, I could set $z=1-1/x$ as a change of variables but I am not sure if this helps.
$2)$ Use the shuffle identities to expand the products
$A. G(a_1, a_2; z) G(a_3; z) $
Attempt: The shuffle product is a linear combination of polylogs of weight $2+1$ in this case which preserves the relative ordering of the indices $a_i$ in each of the terms. So if I understand this correctly, here we simply have $G(a_1, a_2,a_3;z) + G(a_1, a_3, a_2;z) + G(a_3, a_1,a_2;z)$ i.e the only restriciton is that $a_1$ is always to the left of $a_2$, i.e preserving the order. Is that right?
We just need to prove the identities $(4)$ and $(5)$ stated here: $$ \text{Li}_2(1-z)+\text{Li}_2\left(1-\frac{1}{z}\right) = -\frac{1}{2}\log^2 z,\tag{4}$$ $$ \text{Li}_2(z)+\text{Li}_2(1-z)=\frac{\pi^2}{6}-\log z\, \log(1-z)\tag{5}$$ $(5)$ has already been proved in this question.
To prove $(4)$ it is enough to notice that $\text{Li}_2(0)=0$ and: $$ \frac{\log z}{1-z}+\frac{\log z}{z^2-z}=-\frac{\log z}{z},$$ so the values at $z=1$ and the derivatives of the LHS and RHS of $(4)$ match.
By combining $(4)$ and $(5)$ we get: