Coupling fermions with gravity Einstein's gravity does not incorporate the "spinor" nature of fermions. The tetrad formulation or Cartan's theory is suggested as the way to go around this problem - by allowing the spin connection to have torsion. It is known that the spin couples with the torsion.
Please suggest some references where this "spin-torsion coupling" has been discussed in details. Specifically, some concrete examples of calculations where one can see this "coupling" term(s) explicitly.
 A: There is no need to involve torsion in coupling Fermions to gravity.  In the vierbein formalism, there is a well-defined, torsion-free, and metric compatable spin connection in any coordinate patch.  Just use that connection in the Dirac action.  
When  you do this, you must remember that when varying the vierbein in the action functional you must simultaneously  vary the spin connection $\omega_{ij\mu}$ so as to preserve the torsion free condition. This requires 
$$
(\delta \omega_{ij\mu}) e^\mu_k 
=-\frac 12\left\{(\nabla_j \delta e_{ik} -  \nabla_k  \delta e_{ij})
+( \nabla_k \delta e_{ji}- \nabla_i  \delta e_{jk}) -( \nabla_i \delta e_{kj}-\nabla_j \delta e_{ki})\right\},
$$
where 
$$
\delta e_{ij} \equiv   {\bf e}_i\cdot \delta {\bf e}_j= \eta_{ib}[e^{*b}_\alpha \delta e_j^\alpha].
$$
When computing the Stress-Energy tensor, this spin-connection variation gives the additional  terms that convert the canonical but non symmetric Stress energy tensor into the symmetric  Belinfant-Rosen tensor. These extra terms are the contribtion to the energy momentum flux from gradients in the spin density. 
Of course, if you want, you can relax the condition of being torsion free and let $\omega_{ij\mu}$ be  independent of the vierbein ${\bf e}_i$. This leads to the  Einstein-Cartan theory. I personally think that E-C is less pretty than pure Einstein gravity, but E-C is nontheless  a perfectly viable physical theory because the difference in physical predictions between it and pure Einstein are very small. What is not true is the often-seen claim that Dirac requires E-C. It does not.  
For more details see the Wikipedia article on the Belinfante-Rosenfeld tensor, and also the classica paper on Gravitational Anomalies by Ed Witten and Luis Alavarez-Gaume (Nucl.Phys. B234 (1984) 269).
A: A Dirac spinor coupled to gravity (together with Einstein-Hilbert) is described by the action ($c=1$, ($-,+,+,+$) metric signature)
$$
S=\int d^4x~\sqrt{-g}\left[
\frac{1}{16\pi G}R-\bar{\psi}(\gamma^\mu D_\mu+m)\psi
 \right]
$$
where $G$ - Newton's constant, $R$ - scalar curvature, and $\sqrt{-g}\equiv\sqrt{-\det{g_{\mu\nu}}}=\det{e^a_\mu}$ (if you want to use tetrads or metric).
Spin connection ${\omega_\mu}^{ab}$ enters via the covariant derivative 
$$D_\mu\psi=\partial_\mu\psi+\frac{1}{4}{\omega_\mu}^{ab}\gamma_{ab}\psi~,$$ 
here $\gamma_{ab}\equiv\frac{1}{2}(\gamma_a\gamma_b-\gamma_b\gamma_a)$ for gamma matrices. If there is non-zero torsion then you can separate torsion free part ${\tilde{\omega}_\mu}^{ab}$ and contorsion ${K_\mu}^{ab}$:
$${\omega_\mu}^{ab}={\tilde{\omega}_\mu}^{ab}+{K_\mu}^{ab}.$$
If torsion is zero you have only ${\tilde{\omega}_\mu}^{ab}$.
I'll leave it to you to vary the action and find equations of motion - Dirac and Einstein.
Edit: Spin connection is indeed antisymmetric by definition.
