The post has been divided into sections to show some patterns, as well as possible evaluations of,
$$_2F_1\big(s,1-s,1,z\big)$$
with $s = \frac12, \frac13, \frac14, \frac16$ for infinitely many algebraic numbers $z.$
I. Parameter $s=\frac12$
Given the nome $q = e^{\pi i\tau}$ and the Jacobi theta functions $\vartheta_n(0,q)$. Define the modular lambda function,
$$\alpha = \frac{16}{\left(\tfrac{\eta(\tau/2)}{\eta(2\tau)}\right)^8+16} = \left(\frac{\sqrt2\,\eta(\tau/2)\eta^2(2\tau)}{\eta^3(\tau)}\right)^8 = \left(\frac{\vartheta_2(0,q)}{\vartheta_3(0,q)}\right)^4$$
Then we propose for appropriate $\tau$ such as $\tau = \sqrt{-d}$ that the ratios below are algebraic numbers,
\begin{align} \left(\frac{\vartheta_2(0,q)}{\sqrt{_2F_1\big(\tfrac12,\tfrac12,1,\alpha\big)}}\right)^4 &\overset{\color{red}?}=\alpha\\ \left(\frac{\vartheta_4(0,q)}{\sqrt{_2F_1\big(\tfrac12,\tfrac12,1,\alpha\big)}}\right)^4 &\overset{\color{red}?}=1-\alpha\\ \left(\frac{\vartheta_3(0,q)}{\sqrt{_2F_1\big(\tfrac12,\tfrac12,1,\alpha\big)}}\right)^4 &\overset{\color{red}?}=1 \end{align}
Note that adding the first two implies the third. Hence,
$$\big(\vartheta_2(0,q)\big)^4+\big(\vartheta_4(0,q)\big)^4 = \big(\vartheta_3(0,q)\big)^4$$
which is known to be true. As eta quotients in the same order above,
$$\left(\frac{2\eta^2(2\tau)}{\eta(\tau)}\right)^4+\left(\frac{\eta^2\big(\tfrac{\tau}2\big)}{\eta(\tau)}\right)^4 = \left(\frac{\eta^5(\tau)}{\eta^2\big(\tfrac{\tau}2\big)\,\eta^2(2\tau)}\right)^4$$
and the equalities
$$\vartheta_3(0,q) = \sum_{m=-\infty}^\infty q^{m^2} = \frac{\eta^5(\tau)}{\eta^2\big(\tfrac{\tau}2\big)\,\eta^2(2\tau)}$$
has a nice cubic version in the next section.
II. Parameter $s=\frac13$
Given the square of the nome, so $q = e^{2\pi i\tau}$ and the Borwein cubic theta functions $a(q),b(q),c(q)$. Define,
$$\beta = \left(\frac{3}{\left(\tfrac{\eta(\tau/3)}{\eta(3\tau)}\right)^3+3}\right)^3=\left(\frac{c(q)}{a(q)}\right)^3$$
Then we propose,
\begin{align} \left(\frac{c(q)}{_2F_1\big(\tfrac13,\tfrac23,1,\beta\big)}\right)^3 &\overset{\color{red}?}=\beta\\ \left(\frac{b(q)}{_2F_1\big(\tfrac13,\tfrac23,1,\beta\big)}\right)^3 &\overset{\color{red}?}=1-\beta\\ \left(\frac{a(q)}{_2F_1\big(\tfrac13,\tfrac23,1,\beta\big)}\right)^3 &\overset{\color{red}?}=1 \end{align}
Adding the first two implies the third,
$$\big(c(q)\big)^3+\big(b(q)\big)^3=\big(a(q)\big)^3$$
which is also known to be true. As eta quotients,
$$\left(\frac{3\eta^3(3\tau)}{\eta(\tau)}\right)^3+\left(\frac{\eta^3(\tau)}{\eta(3\tau)}\right)^3 =\left(\frac{\eta^3(\tau)+9\eta^3(9\tau)}{\eta(3\tau)}\right)^3$$
For the square of the nome, so $q = e^{2\pi i\tau}$, then the cubic analogue,
$$a(q) = \sum_{m,n\,=-\infty}^\infty q^{m^2+mn+n^2} = \frac{\eta^3(\tau)+9\eta^3(9\tau)}{\eta(3\tau)}$$
III. Parameter $s=\frac14$
Given the square of the nome $q = e^{2\pi i\tau}$. Define, $$\gamma = \left(\frac{8}{\left(\tfrac{\eta(\tau/2)}{\eta(2\tau)}\right)^8+8}\right)^2 = \left(\frac{C(q)}{A(q)}\right)^2$$
Then,
\begin{align} \left(\frac{C(q)}{_2F_1\big(\tfrac14,\tfrac34,1,\gamma\big)^2}\right)^2 &\overset{\color{red}?}=\gamma\\ \left(\frac{B(q)}{_2F_1\big(\tfrac14,\tfrac34,1,\gamma\big)^2}\right)^2 &\overset{\color{red}?}=1-\gamma\\ \left(\frac{A(q)}{_2F_1\big(\tfrac14,\tfrac34,1,\gamma\big)^2}\right)^2 &\overset{\color{red}?}=1 \end{align}
Again, the first two implies the third,
$$\big(C(q)\big)^2+\big(B(q)\big)^2=\big(A(q)\big)^2$$
where $C(q), B(q), A(q)$ are defined by the eta quotients,
$$\left(\frac{8\eta^8(2\tau)}{\eta^4(\tau)}\right)^2+\left(\frac{\eta^8(\tau)}{\eta^4(2\tau)}\right)^2=\left(\frac{\eta^8(\tau)+32\eta^8(4\tau)}{\eta^4(2\tau)}\right)^2$$
Unlike $a(q)$, I am not aware of a sum for $A(q)$,
$$A(q) = \sum_{m=-\infty}^\infty ??\, = \frac{\eta^8(\tau)+32\eta^8(4\tau)}{\eta^4(2\tau)}$$
But note that,
$$\frac{24}{A(q)-1}= \frac1{q} -1 - 3q + 6q^2 + q^3 - 20q^4 + 24q^5 + 38q^6 - 132q^7 + \dots$$
which seems to be A335227. Update: Michael Somos pointed out that,
$$A(q) = 1 + 24q + 24q^2 + 96q^3 + 24q^4 + 144q^5 + 96q^6 + 192q^7 + \dots$$
which is A004011 and is the theta series of $D_4$ lattice. So we finally have a sum,
\begin{align}A(q) &= 1+24\sum_{n=1}^\infty\frac{n q^n}{1+q^n}\\[4pt] &=\big(\vartheta_2(0,q)\big)^4+\big(\vartheta_3(0,q))^4\qquad \end{align}
related to the odd divisor function, and where all $q$ are $q=e^{2\pi i\tau}.$
IV. Parameter $s=\frac16$
Given the golden ratio $\phi$, then,
$$_2F_1\left(\frac16,\frac56,1,\frac{\phi^{-5}}{5\sqrt5}\right) = \left(\frac{3\sqrt3}{8}\right)^{1/3}\left(\frac{5\sqrt5}{8}\right)^{1/6}\,\frac{\Gamma\big(\tfrac13\big)^3}{2\pi^2}$$
(Note: The rest of the section has been moved to a MSE post to trim this post.)
V. Context
These observations arose from evaluations of the complete elliptic integral of the first kind, $K(k)$. For ex., given the tribonacci constant $T$, the real root of $T^3-T^2-T-1=0$, then,
$$K(k_{11}) = \frac{\pi\,(2T)^{2/3}}{2\;}\times \frac{\Gamma\big(\tfrac1{11}\big) \Gamma\big(\tfrac3{11}\big) \Gamma\big(\tfrac4{11}\big) \Gamma\big(\tfrac5{11}\big) \Gamma\big(\tfrac9{11}\big)}{11^{1/4}(2\pi)^3}$$
Manipulating the $s=\frac12$ relations above, we can have a much shorter version,
$$K(k_{11}) = \frac{\pi\,(2T)^{4/3}}{2\;}\times\eta^2\big(\sqrt{-11}\big)\quad\quad$$
Equating the two formulas, this also gives the explicit evaluation of $\eta(\tau)$.
VI. Question
Q: So are the proposed relations $M\overset{\color{red}?}=N$ with the red question marks in fact true?
