All identities except possibly $(8)$ can be proved elegantly using Wilf-Zeilberger pairs.
To be precise, if $F(n,k), G(n,k)$ are two $\mathbb{C}$-valued functions satisfying $$\tag{*}F(n+1,k) - F(n,k) = G(n,k+1)-G(n,k)$$ then via some telescoping, one has, if $\lim_{n\to \infty} G(n,k) = 0$ for each $k\geq 0$, then $$\sum_{k\geq 0} F(0,k) = \lim_{n\to\infty} \sum_{k\geq 0} F(n,k) + \sum_{n\geq 0} G(0,n)$$
If one takes $$F(n,k) = \frac{\Gamma (-a+c+n+1) \Gamma \left(a+k+n+\frac{1}{2}\right) \Gamma (-b+c+n+1) \Gamma \left(b+k+n+\frac{1}{2}\right)}{\Gamma (a-n) \Gamma (b-n) \Gamma \left(k+2 n+\frac{3}{2}\right) \Gamma \left(c+k+2 n+\frac{3}{2}\right) \Gamma (-a-b+c+2 n+1)}$$ then one find its WZ-mate $G(n,k)$ with $\lim_{n\to\infty}\sum_{k\geq 0}F(n,k) = 0$, and we have $$\sum_{k\geq 0}\frac{32 \left(a+d+\frac{1}{2}\right)_k \left(b+d+\frac{1}{2}\right)_k}{(2 d+2 k+1) (2 c+2 d+2 k+1) \left(d+\frac{1}{2}\right)_k \left(c+d+\frac{1}{2}\right)_k} \\ = \sum_{n\geq 1}\frac{(1-a)_{n-1} (1-b)_{n-1} (-a+c+1)_{n-1} \left(a+d+\frac{1}{2}\right)_{n-1} (-b+c+1)_{n-1} \left(b+d+\frac{1}{2}\right)_{n-1} \times P}{\left(d+\frac{1}{2}\right)_{2 n} \left(c+d+\frac{1}{2}\right)_{2 n} (-a-b+c+1)_{2 n}}$$
here $P$ is a long polynomial that pops up in computing WZ-mate $G(n,k)$: $8 a^2 b^2-8 a^2 b c-16 a^2 b n+8 a^2 c n+8 a^2 n^2-8 a b^2 c-16 a b^2 n+8 a b c^2+8 a b c d+48 a b c n-4 a b c+8 a b d^2+32 a b d n-8 a b d+64 a b n^2-16 a b n+2 a b-8 a c^2 d-24 a c^2 n+4 a c^2-16 a c d^2-72 a c d n+24 a c d-104 a c n^2+52 a c n-8 a c-8 a d^3-56 a d^2 n+20 a d^2-128 a d n^2+88 a d n-14 a d-112 a n^3+96 a n^2-30 a n+3 a+8 b^2 c n+8 b^2 n^2-8 b c^2 d-24 b c^2 n+4 b c^2-16 b c d^2-72 b c d n+24 b c d-104 b c n^2+52 b c n-8 b c-8 b d^3-56 b d^2 n+20 b d^2-128 b d n^2+88 b d n-14 b d-112 b n^3+96 b n^2-30 b n+3 b+8 c^3 d+16 c^3 n-4 c^3+16 c^2 d^2+80 c^2 d n-24 c^2 d+104 c^2 n^2-56 c^2 n+8 c^2+8 c d^3+80 c d^2 n-20 c d^2+232 c d n^2-128 c d n+14 c d+224 c n^3-180 c n^2+44 c n-3 c+16 d^3 n+104 d^2 n^2-40 d^2 n+224 d n^3-168 d n^2+28 d n+168 n^4-176 n^3+58 n^2-6 n$
When $(a,b,c,d) = (0,0,0,0)$, above equality reduces to $$\frac{\pi^2}{2} = \sum_{k\geq 0} \frac{4}{(1+2k)^2} = \sum_{n\geq 0} 64^{-n} \frac{(1/2)_n (1)_n^3}{(1/4)_n^2 (3/4)_n^2} \frac{3-23n+42n^2}{n^3(2n-1)}$$ which is exactly $(2)$. Those involving harmonic number $(9),(10),(11),(12)$ can be proved by comparing coefficient in Taylor expansion of above equality around $(a,b,c,d) = (0,0,0,0)$, coefficients of $\sum_{k\geq 0}$ will give the so-called multiple $t$-values (special case of alternating multiple zeta values), whose values, when weight is small, are well-understood.
For $(1)$ and $(4)$-$(7)$, one repeats the above process, except now using another $F(n,k)$: $F(n,k) = \frac{\Gamma (-a+c+2 n+1) \Gamma (a+k+n+1) \Gamma (-b+c+2 n+1) \Gamma (b+k+n+1)}{\Gamma (a-n) \Gamma (b-n) \Gamma (k+2 n+2) \Gamma (c+k+3 n+2) \Gamma (-a-b+c+3 n+1)}$
Unfortunately, one cannot deduce $(8)$ in this way.
For $(3)$ and $(13)$-$(15)$, one repeats the above process, with $F(n,k)$: $F(n,k) = \frac{\Gamma (-a+c+n+1) \Gamma (a+k+2 n+1) \Gamma (-b+c+n+1) \Gamma (b+k+2 n+1)}{\Gamma \left(a+\frac{1}{2}\right) \Gamma \left(b+\frac{1}{2}\right) \Gamma \left(k+2 n+\frac{3}{2}\right) \Gamma (c+k+3 n+2) \Gamma \left(-a-b+c+n+\frac{1}{2}\right)}$