So in the process of trying to find a derivation for this answer, the following interesting equalities arose (one can check with Wolfram Alpha/Mathematica):
$$\frac{8\sqrt{2}G^4}{5\pi^2} \left(\left(7 \sqrt{2}-10\right) \beta +5 \left(\sqrt{2}-2\right)\right) = -\pi/2,\tag1$$
$$-\frac{4}{3} \left(\alpha\left(\sqrt{2}-1\right)^2 +6 \ln \left(\sqrt{2}-1\right)\right) = 7\ln 2 - \ln(17-12\sqrt{2})-\pi/2,\tag2$$
where $G = \Gamma\left(\frac{3}{4}\right)$, $\alpha = {}_3F_2\left(1,1,\frac{5}{4};\frac{7}{4},2;3-2\sqrt{2}\right)$ and $\beta = {}_3F_2\left(1,\frac{3}{2},\frac{7}{4};\frac{9}{4},\frac{5}{2};3-2\sqrt{2}\right)$.
Doing some simplification and solving will tell you:
$${}_3F_2\left(1,1,\frac{5}{4};\frac{7}{4},2;3-2\sqrt{2}\right) = \frac{3}{4}\cdot\frac{\pi/2-7\ln{2}-4\ln(\sqrt{2}-1)}{(\sqrt{2}-1)^2},\tag4$$
$${}_3F_2\left(1,\frac{3}{2},\frac{7}{4};\frac{9}{4},\frac{5}{2};3-2\sqrt{2}\right) = \frac{5}{4}\cdot\frac{(\pi/2)^3-4(\sqrt{2}-1)G^4}{G^4(\sqrt{2}-1)^3}.\tag5$$
It's very bizarre that aside from the Gamma function valued at 3/4, these come out to (relatively) nice closed forms. My guess is that they are a result of integrals, but I have no idea what those integrals could be. Mathematica doesn't get anywhere with the hypergeometric functions, so I'm a bit stuck.
Note: $(\sqrt{2}-1)^2 = 3 - 2\sqrt{2}$. That last number seems to come out, sort of, in the result of the hypergeometric functions' closed form, but I can't place why.
EDIT: Here are some ways of describing both functions simultaneously, which may help in some way:
${}_3F_2\left(a,b,c;c+\frac{1}{2},b+1;z\right)$, ${}_3F_2\left(a,b,c;c+\frac{1}{2},a+b;z\right)$, ${}_3F_2\left(a,b,b+\frac{1}{4};a+b-\frac{1}{4},a+b;z\right)$, ${}_3F_2\left(1,a,a+\frac{1}{4};a+\frac{3}{4},a+1;z\right)$