On $_2F_1(\tfrac13,\tfrac23;\tfrac56;\tfrac{27}{32}) = \tfrac85$ and $_2F_1(\tfrac14,\tfrac34;\tfrac78;\tfrac{48}{49}) = \tfrac{\sqrt7}3(1+\sqrt2)$
Solution 1:
This is a partial answer to the 2nd question. It explains how to find "isolated" examples of hypergeometric functions that have algebraic values at algebraic points using educated guess combined with numerical experimentation. The following example comes from the transformation
$\tag{1}$
taken from this article:
$$
{}_2F_1\left(\frac{3}{10},\frac{2}{5};\frac{9}{10};\frac{1}{\phi ^2}\right)=\frac{\phi ^{9/5}}{\sqrt{5}}.\tag{2}
$$
Suppose one evaluates the RHS of (1) at the point $z=1/\phi$ ($\phi-$ Golden ratio), where the argument of the hypergeometric function on the LHS of (1) vanishes due to $1-1/\phi-1/\phi^2=0$. One might expect that ${}_2F_1\left(\frac{3}{10},\frac{2}{5};\frac{9}{10};\frac{1}{\phi ^2}\right)$ equals $\frac{1}{\sqrt[20]{1-1/\phi^2}} \frac{1}{\sqrt[4]{1+4/ \phi-1/\phi^2}}$. However numerical evaluation shows that this is not the case, and this is due to the fact that $z=1/\phi$ does not belong to the validity range of (1). But this doesn't matter because one might easily find possible closed form (2) for ${}_2F_1\left(\frac{3}{10},\frac{2}{5};\frac{9}{10};\frac{1}{\phi ^2}\right)$ by numerical experimentation. After application of Euler transformation (2) takes the nice form
$${}_2F_1\left(\frac{1}{2},\frac{3}{5};\frac{9}{10};\frac{1}{\phi ^2}\right)=\frac{\phi ^{2}}{\sqrt{5}},\tag{3}$$
where parameters of the hypergeometric function satisfy $c=1+a-b$.
I want to stress that (2) and (3) are only conjectures because this method doesn't provide the proof.
To obtain another algebraic value consider the transformation (22) in the article https://arxiv.org/abs/0807.4808 $$ \, _2F_1\left(a,\frac{2 a+1}{6};\frac{1}{2};x\right)=\left(\frac{x}{3}+1\right)^{-a}\, _2F_1\left(\frac{a}{3},\frac{a+1}{3};\frac{1}{2};\frac{x (9-x)^2}{(x+3)^3}\right). $$ $x=9$ does not belong to validity range of (1) but one expects that ${}_2F_1\left(a,\frac{2 a+1}{6};\frac{1}{2};9\right)$ is algebraic and indeed according to DLMF 15.8.2 and DLMF 15.4.32 we have $$ {}_2F_1\left(a,\frac{2 a+1}{6};\frac{1}{2};9\right)=\frac{e^{-\pi i a} \cos \pi a-e^{-\pi i/6-\pi ia/3}\sin \pi a}{2^{2 a+1} \sin \left(\frac{\pi }{6}-\frac{2 \pi a}{3}\right)}. $$