I'm going to only cover conjecture $(4)$. We can use elliptic functions to show it is true.
To simplify setup, we will adopt all conventions and notation as in
this answer.
For $x \in (0, 1)$, let $F(x) = x _3F_2(1,1,\frac54;2,\frac74;x)$, we have
$$
F(x) = \sum_{k=0}^\infty \frac{x^{k+1}}{k+1}\frac{(\frac54)_k}{(\frac74)_k}
= \frac{\Gamma(\frac74)}{\Gamma(\frac12)\Gamma(\frac54)}
\sum_{k=0}^\infty \frac{x^{k+1}}{k+1}\int_0^1 t^{\frac14+k}(1-t)^{-\frac12}dt\\
= -\frac{3}{2\sqrt{2}\omega}\int_0^1 \log(1-xt) t^{-\frac34}(1-t)^{-\frac12} dt
$$
Let $x = \alpha^2$, $\beta = \frac{1-\alpha}{1+\alpha}$ and substitute $t$ by $\left(\frac{p-\frac12}{p + \frac12}\right)^2$ and then $p$ by $\wp(z)$, we have:
$$\begin{align}
F(\alpha^2)
&= -\frac{3}{\omega}\int_{\infty}^{\frac12} \log\left[(1-\alpha^2)\frac{(p+\frac12\beta)(p+\frac12\beta^{-1})}{(p+\frac12)^2}\right] \frac{dp}{\sqrt{4p^3-p}}\\
&= -\frac{3}{2\omega}\int_{-\omega}^\omega \log\left[(1-\alpha^2)\frac{(\wp(z)+\frac12\beta)(\wp(z)+\frac12\beta^{-1})}{(\wp(z)+\frac12)^2}\right] dz
\end{align}\tag{*1}
$$
Notice $1 - (\sqrt{2}-1)^2 = 2 (\sqrt{2}-1)$, conjecture $(4)$ can be rewritten as
$$F((\sqrt{2}-1)^2) \stackrel{?}{=}\frac34\left[ \frac{\pi}{2} - 3\log 2\right] - 3 \log\left[1 - (\sqrt{2}-1)^2\right]$$
Compare this with $(*1)$, we find conjecture $(4)$ is equivalent to
$$\frac{1}{2\omega}\int_{-\omega}^\omega
\log\left[\frac{(\wp(z)+\frac12\beta)(\wp(z)+\frac12\beta^{-1})}{(\wp(z)+\frac12)^2}\right] dz \stackrel{?}{=}\frac{1}{4}(3\log 2 - \frac{\pi}{2})\tag{*2}
$$
when $\alpha = \beta = \sqrt{2}-1$.
Like the other answer, we can express the RHS of $(*2)$ using Weierstrass sigma function. Before we do that, I will like to point out
$$\left(\frac{d}{dz}\wp(z)\right) ^2 = 4 \wp(z)^3 - \wp(z)^3
\;\;\implies\;\;
\left(\frac{d}{dz}\frac{1}{4\wp(iz)}\right)^2 = 4 \left(\frac{1}{4\wp(iz)}\right)^3 -
\left(\frac{1}{4\wp(iz)}\right)
$$
One consequence of this is if we pick a $\rho$ such that $\wp( \pm ( \omega' + \rho) ) = -\frac12\beta$, then
$$\wp(\pm ( \omega' + i\rho)) =
\wp(\pm( \omega' - i\rho)) = -\frac12\beta^{-1}
\quad\text{ and }\quad\wp( \pm( \omega' - \rho)) = -\frac12\beta$$
When $\alpha = \beta = \sqrt{2}-1$, we can pick $\rho = \frac{\omega}{2}$. The RHS of $(*2)$ becomes
$$\frac{1}{2\omega}\int_{-\omega}^{\omega}
\left\{
\sum_{k=0}^3\log\left[
\frac{\sigma(z + \omega' + i^k\rho)}{\sigma(\omega' +i^k\rho))}
\right] - 4 \log\left[
\frac{\sigma(z + \omega')}{\sigma(\omega')}
\right] \right\} dz$$
One can perform same analysis as in my other answer by playing with $\varphi_{\pm}(\tau)$ functions. Because of the symmetry of $\wp(z)$ around the point $z = \omega' = i\omega$, their contributions will cancel out with each other. The net result is
$$\text{RHS}[*2] = \log\left[\frac{\sigma(\omega')^4}{\prod_{k=0}^3\sigma(\omega' + i^k\rho))}\right]\tag{*3}$$
For Weierstrass elliptic functions with general $g_2, g_3$, we know
$$\begin{cases}
\wp'(z) &= -\frac{\sigma(2z)}{\sigma(z)^4}\\
\sigma(z+2\omega) &= -e^{2\eta(z+\omega)}\sigma(z)\\
\sigma(z+2\omega') &= -e^{2\eta'(z+\omega')}\sigma(z)
\end{cases}$$
This implies
$$\sigma(z+\omega)^4 = -\frac{\sigma(2z+2\omega)}{\wp'(z+\omega)} = e^{2\eta(2z+\omega)}\frac{\sigma(2z)}{\wp'(z+\omega)} \quad\implies\quad
\sigma(\omega)^4 = e^{2\eta\omega}\frac{2}{\wp''(\omega)}$$
For our case where $(g_2,g_3) = (1,0)$, we know $\eta = \frac{\pi}{4\omega}$, this leads to $\sigma(\omega) = e^{\frac{\pi}{8}}\sqrt[4]{2}$. Let me call this number $\Omega$.
When $g_3 = 0$, the double poles of $\wp(z)$ forms a square lattice. This 4-fold symmetry
around the origin give us $\sigma(i\omega) = i\sigma(\omega) = i\Omega$.
The value of sigma functions at other points in $(*3)$ can be deduced in similar manner:
$$\begin{align}
\left|\sigma\left(\frac{\omega'}{2}\right)\right|
&= \left|\frac{\sigma(\omega')}{\wp'(\frac{\omega'}{2})}\right|^{1/4} = \left(\frac{\Omega}{\sqrt{2}+1}\right)^{1/4}\\
\left|\sigma\left(\frac{3\omega'}{2}\right)\right|
&= \left|e^{\eta'\omega'}\sigma\left(-\frac{\omega'}{2}\right)\right| = e^{\pi/4}\left(\frac{\Omega}{\sqrt{2}+1}\right)^{1/4}\\
\left|\sigma\left(\omega'\pm\frac{\omega}{2}\right)\right|
&= \left|\frac{\sigma\left(2\omega'\pm\omega\right)}{\wp'(\omega'\pm\frac{\omega}{2})}\right|^{1/4}
= \left|e^{2\eta'(\omega'\pm\omega)}\frac{\Omega}{\sqrt{2}-1}\right|^{1/4}
= e^{\pi/8}\left(\frac{\Omega}{\sqrt{2}-1}\right)^{1/4}
\end{align}$$
Combine all this, we find
$$\text{RHS}[*2] = \log\left(\frac{\Omega^4}{e^{\pi/2}\Omega}\right)
= 3\log\Omega - \frac{\pi}{2} = 3\left(\frac14\log 2 + \frac{\pi}{8}\right) - \frac{\pi}{2} = \frac14\left(3\log 2 - \frac{\pi}{2}\right)$$
i.e Conjecture $(4)$ is true.
Update
It turns out there is a cleaner algebraic relation between the hypergeometric function
in conjecture $(4)$ and elliptic functions. Using the addition formula for sigma function:
$$\wp(z)-\wp(u) = - \frac{\sigma(z+u)\sigma(z-u)}{\sigma(z)^2\sigma(u)^2}$$
We find
$$\frac{\sigma(\omega')^4}{\prod_{k=0}^3\sigma(\omega'+ i^k\rho)}
=\left(\frac{1}{\sigma(\rho)^2(\wp(\omega')-\wp(\rho))}\right)
\left(\frac{1}{\sigma(i\rho)^2(\wp(\omega')-\wp(i\rho))}\right)
$$
Notice
$$\begin{cases}
\wp(\omega'\pm\rho) &= -\frac12\beta\\
\wp(\omega'\pm i\rho) &= -\frac12\beta^{-1}
\end{cases}
\quad\implies\quad
\begin{cases}
\wp(\pm \rho) &= \frac{1}{2\alpha}\\
\wp(\pm i\rho) &= -\frac{1}{2\alpha}\\
\end{cases}
$$
We get
$$\frac{\sigma(\omega')^4}{\prod_{k=0}^3\sigma(\omega'+ i^k\rho)} =
\frac{4\alpha^2}{\sigma(\rho)^4(1-\alpha^2)}
$$
Which in terms implies a simple relation between $F(\alpha^2)$ and $\sigma(\rho)$:
$$_3F_2\left(1,1,\frac54; 2,\frac74; \alpha^2\right) = \frac{F(\alpha^2)}{\alpha^2} = \frac{12}{\alpha^2}\log\left(\frac{\sigma(\rho)}{\sqrt{2\alpha}}\right)$$
When $\alpha = \sqrt{2}-1$, $\rho = \frac{\omega}{2}$ and since
$\sigma(\frac{\omega}{2}) = \left(\frac{\Omega}{\sqrt{2}+1}\right)^{1/4}$, we again
obtain
$$_3F_2\left(1,1,\frac54; 2,\frac74; (\sqrt{2}-1)^2\right)
= \frac{3}{(\sqrt{2}-1)^2}\log\left(\frac{e^{\pi/8}\sqrt[4]{2}}{(\sqrt{2}+1)4(\sqrt{2}-1)^2}\right)
= \frac{3}{4(\sqrt{2}-1)^2}\left[\frac{\pi}{2} - 7\log 2 - 4\log(\sqrt{2}-1)\right]
$$
Setting $\rho$ to other rational multiples of $\omega$, we can deduce a bunch of similar identities. For example,
- When $\rho = \omega$, it is clear $\alpha = 1$ and $\sigma(\omega) = \Omega$, this leads to
$$
_3F_2\left(1,1,\frac54; 2,\frac74; 1\right) = 12\log\left(\frac{\Omega}{\sqrt{2}}\right)
= \frac32\left(\pi- 2\log 2\right)
$$
- When $\rho = \frac{2\omega}{3}$, one can use the triplication formula of sigma function
$$\frac{\sigma(3z)}{\sigma(z)^9} = 3\wp(z)\wp'(z)^2 - \frac14 \wp''(z)^2$$
to show
$$\alpha = \sqrt{2\sqrt{3}-3}
\quad\text{ and }\quad
\sigma\left(\frac{2\omega}{3}\right) = e^{\pi/18} 3^{1/8} (2-\sqrt{3})^{1/12}$$
This leads to the identity
$$_3F_2\left(1,1,\frac54; 2,\frac74; 2\sqrt{3}-3\right) =
\frac{1}{2\sqrt{3}-3}\left(\frac{2\pi}{3} - 6\log 2 - 2\log(2-\sqrt{3})\right)
$$
