Closed form of $\operatorname{Li}_2(\varphi)$ and $\operatorname{Li}_2(\varphi-1)$

The most curious and important dilogarithm identity is the pentagonal one: \begin{align} \mathrm{Li}_2\left(\frac{x}{1-y}\right)+\mathrm{Li}_2\left(\frac{y}{1-x}\right)-\mathrm{Li}_2(x)-\mathrm{Li}_2(y)-\mathrm{Li}_2\left(\frac{xy}{(1-x)(1-y)}\right)=\\ =\ln(1-x)\ln(1-y).\tag{1} \end{align} Denote $\alpha=\frac{\sqrt{5}-1}{2}$. Then it is very easy to check that \begin{align} \left(\frac{x}{1-y},\frac{y}{1-x},x,y,\frac{xy}{(1-x)(1-y)}\right)_{x=y=1-\alpha}&=\left(\alpha,\alpha,1-\alpha,1-\alpha,1-\alpha\right).\tag{2} \end{align} On the other hand, one has $$\mathrm{Li}_2(x)+\mathrm{Li}_2(1-x)=\frac{\pi^2}{6}-\ln x\ln(1-x).\tag{3}$$ Combining (1), (2) and (3), we can express $\mathrm{Li}_2(\alpha)$ and $\mathrm{Li}_2(1-\alpha)$ in terms of elementary functions. These are basic special values from which one can deduce the others.

Now concerning the values you are interested in. The first of your formulas is meaningless as written, since the dilogarithm has branch cut $[1,\infty)$. The second is almost okay, if you correct the sign in front of $\Phi$ on the right. To obtain the corresponding evaluation, it suffices to use the identity $$\mathrm{Li}_2(x)+\mathrm{Li}_2(-x)-\frac12\mathrm{Li}_2(x^{2})=0\tag{4}$$ with $x=\alpha$ and the fact that $\alpha^2=1-\alpha$.

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Addition (for completeness): Denoting $$A=\mathrm{Li}_2(\alpha),\qquad B=\mathrm{Li}_2(1-\alpha),\qquad C=\mathrm{Li}_2(-\alpha),$$ we have from the equations (1)-(4): \begin{align} &2A-3B=\ln^2\alpha,\\ &A+B=\frac{\pi^2}{6}-2\ln^2\alpha,\\ &A+C-\frac12 B=0. \end{align} The solution of this system is given by \begin{align} &A=\frac{\pi^2}{10}-\ln^2\alpha,\\ &B=\frac{\pi^2}{15}-\ln^2\alpha,\\ &C=-\frac{\pi^2}{15}+\frac12\ln^2\alpha. \end{align}