Complete this table of general formulas for algebraic numbers $u,v$ and $_2F_1\big(a,b;c;u) =v $?

(This extends this post.) Given fixed rationals $a,b,c,$ the problem of determining,
$$_2F_1\big(a,b;c;u) =v $$
such that both $u,v$ are algebraic numbers may be solved by appealing to modular functions/forms like the j-function $j(\tau)$ and Dedekind eta function $\eta(\tau)$.

In the tables below, I derived the formulas empirically using data from Zucker and Joyce in “Special values of the hypergeometric series II, III”. However, I’m missing five examples and their formulas.

I. Type $a+b=c=\color{blue}{\tfrac12}$.

$$\begin{aligned}
&\,_2F_1\big(\tfrac14,\tfrac14;\tfrac12;(1-2\alpha_1)^2\big),\quad \frac1{\alpha_1}-1=\frac1{16}\Big(\tfrac{\eta(\tau/4)}{\eta(\tau)}\Big)^8,\quad \tau_1=N\sqrt{-4}\\[2mm]
&\,_2F_1\big(\tfrac16,\tfrac13;\tfrac12;(1-2\alpha_2)^2\big),\quad \frac1{\alpha_2}-1=\frac1{27}\Big(\tfrac{\eta(\tau/3)}{\eta(\tau)}\Big)^{12},\quad \tau_2=N\sqrt{-3}\\[2mm]
&\,_2F_1\big(\tfrac18,\tfrac38;\tfrac12;(1-2\alpha_3)^2\big),\quad \frac1{\alpha_3}-1=\frac1{64}\Big(\tfrac{\eta(\tau/2)}{\eta(\tau)}\Big)^{24},\quad \tau_3=N\sqrt{-2}\\[2mm]
&\,_2F_1\big(\tfrac1{12},\tfrac5{12};\tfrac12;(1-2\alpha_4)^2\big),\quad \frac{1}{\alpha_4(1-\alpha_4)}=\frac1{432}\,j(\tau),\quad\tau_4=N\sqrt{-1}\end{aligned}$$

though the argument $z_4$ of the fourth can be found more simply as $z_4 = (1-2\alpha_4)^2 = \frac{j(\tau)-1728}{j(\tau)}$.
Examples:
$$_2F_1\left(\tfrac14,\tfrac14;\tfrac12;\,9(11-8\sqrt2)^2\right)=\tfrac{3}{4\sqrt2}(1+\sqrt2)$$
$$_2F_1\left(\tfrac16,\tfrac13;\tfrac12;\,\tfrac{25}{27}\right)=\tfrac{3\sqrt3}{4}$$
$$_2F_1\left(\tfrac18,\tfrac38;\tfrac12;\tfrac{2400}{2401}\right)=\tfrac{2\sqrt7}{3}$$
$$_2F_1\left(\tfrac1{12},\tfrac5{12};\tfrac12;\tfrac{1323}{1331}\right)=\tfrac{3\,\sqrt[4]{11}}{4}$$
using $\tau_1=2\sqrt{-4},\;\tau_2=2\sqrt{-3},\;\tau_3=3\sqrt{-2},\;\tau_4=2\sqrt{-1}$, respectively. And so on for any integer $N>1$.

II. Type $a+b=c=\color{blue}{\tfrac23}$.

$$\begin{aligned}
&\,_2F_1\big(\tfrac14,\tfrac5{12};\tfrac23;\beta_1\big),\quad \color{red}{\beta_1 =\,?} \\[2mm]
&\,_2F_1\big(\tfrac16,\tfrac12;\tfrac23;\beta_2\big),\quad \frac{4\beta_2}{(1-\beta_2)^2}=\frac{-j(\tau)}{12^3}\\[2mm]
&\,_2F_1\big(\tfrac18,\tfrac{13}{24};\tfrac23;\beta_3\big),\quad \color{red}{\beta_3 =\,?} \\[2mm]
&\,_2F_1\big(\tfrac1{12},\tfrac7{12};\tfrac23;\beta_4\big),\quad\frac{\beta_4}{1-\beta_4}=\frac{-j(\tau)}{12^3}\end{aligned}$$

Examples: Use $\tau = \frac{1+N\sqrt{-3}}2$, like $\tau = \frac{1+3\sqrt{-3}}2$,

$$_2F_1\big(\tfrac16,\tfrac12;\tfrac23;\tfrac{125}{128}\big) =\tfrac43\,2^{1/6}$$ $$_2F_1\big(\tfrac1{12},\tfrac7{12};\tfrac23;\tfrac{64000}{64009}\big) =\tfrac23\,253^{1/6}$$

III. Type $a+b=c=\color{blue}{\tfrac34}$.

$$\begin{aligned}
&\,_2F_1\big(\tfrac14,\tfrac12;\tfrac34;(1-2\gamma_1)^2\big), \quad\frac1{\gamma_1}-1=\frac18\Big(\tfrac{\sqrt2\,\eta(2\tau)}{\zeta_{48}\,\eta(\tau)}\Big)^{12}\\[2mm]
&\,_2F_1\big(\tfrac16,\tfrac7{12};\tfrac34;(1-2\gamma_2)^2\big),\quad\color{red}{\gamma_2 =\,?} \\[2mm]
&\,_2F_1\big(\tfrac18,\tfrac58;\tfrac34;(1-2\gamma_3)^2\big),\quad\frac1{\gamma_3}-1=\frac1{64}\Big(\tfrac{\sqrt2\,\eta(2\tau)}{\zeta_{48}\,\eta(\tau)}\Big)^{24}\\[2mm]
&\,_2F_1\big(\tfrac1{12},\tfrac23;\tfrac34;(1-2\gamma_4)^2\big),\quad\color{red}{\gamma_4 =\,?}
\end{aligned}$$
where $\zeta_{48} = e^{2\pi i/48}$.

Examples: Use $\tau = \frac{1+N\sqrt{-1}}2$, like $\tau = \frac{1+5\sqrt{-1}}2$,

$$_2F_1\big(\tfrac14,\tfrac12;\tfrac34;\tfrac{80}{81}\big)=\tfrac95$$
$$_2F_1\big(\tfrac18,\tfrac58;\tfrac34;\tfrac{25920}{25921}\big)=\tfrac35\,161^{1/4}$$

IV. Type $a+b=c=\color{blue}{\tfrac56}$.

$$\begin{aligned}
&\,_2F_1\big(\tfrac13,\tfrac12;\tfrac56;(1-2\delta_2)^2\big),\quad\;\frac1{\delta_2}-1=\frac1{\sqrt{27}}\left(\tfrac{\eta\big(\frac{\tau+1}{3}\big)}{\eta(\tau)}\right)^{6}\\[2mm]
&\,_2F_1\big(\tfrac14,\tfrac7{12};\tfrac56;(1-2\delta_3)^2\big),\quad\color{red}{\delta_3 =\,?} \\[2mm]
&\,_2F_1\big(\tfrac16,\tfrac23;\tfrac56;(1-2\delta_4)^2\big),\quad\;\frac1{\delta_4}-1=\frac1{27}\left(\tfrac{\eta\big(\frac{\tau+1}{3}\big)}{\eta(\tau)}\right)^{12}
\end{aligned}$$
Note: Of course, $_2F_1\big(\tfrac12,\tfrac13;\tfrac56;z\big) =\,_2F_1\big(\tfrac13,\tfrac12;\tfrac56;z\big)\,$ so the first form is superfluous.

Examples: Use $\tau = \frac{1+N\sqrt{-3}}2$, like $\tau = \frac{1+5\sqrt{-3}}2$,

$$\begin{aligned}
&\,_2F_1\big(\tfrac13,\tfrac12;\tfrac56;\tfrac45\big)=\;\tfrac35\sqrt5 \\[2mm]
&\,_2F_1\big(\tfrac16,\tfrac23;\tfrac56;\tfrac{80}{81}\big)=\tfrac35\,(9\sqrt5)^{1/3}\end{aligned}$$
with the last one discussed in this post.


Q: How do we find the five missing formulas (?) for the variables in red?

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