Your spinors are real and your Dirac matrices in the Majorana representation are imaginary, so

• $\gamma^0$ is Hermitean A(ntisymmetric)
• $\gamma^i$ are Antihermitean S(ymmetric)
• $\gamma^5$ is Hermitean A(ntisymmetric) $\leadsto$ check this from the above!

Thus $$ \bar{\epsilon}\eta= \epsilon^T \gamma^0 \eta =-\eta^T \gamma^{0~~T} \epsilon =\eta^T \gamma^{0 }\epsilon = \bar{\eta} \epsilon , $$ So now you got your baseline. Make sure you appreciate every step.

The rest follow trivially from the above properties of the γ matrices and the above properties, so supplanting $\gamma^0$, you now have $$(\gamma^0 \gamma^5)^T= \gamma^5 \gamma^0= -\gamma^0 \gamma^5,$$ just like for the scalar, so $\bar{\epsilon}\gamma_{5}\eta=\bar{\eta}\gamma_{5}\epsilon$. (Anti-)likewise, $$ (\gamma^0 \gamma^\mu)^T= \gamma^0 \gamma^\mu, $$ where you can marvel at how the time-like and space like parts synchronize to yield a uniform answer, hence $\bar{\epsilon}\gamma^{\mu}\eta=-\bar{\eta}\gamma^{\mu}\epsilon$. Finally, for the axial, $\gamma^5$ flips the sign to yield $\bar{\epsilon}\gamma^{\mu}\gamma_{5}\eta=\bar{\eta}\gamma^{\mu}\gamma_{5}\epsilon$ .

Your spinors are real and your Dirac matrices in the Majorana representation are imaginary, so

• $\gamma^0$ is Hermitean A(ntisymmetric)
• $\gamma^i$ are Antihermitean S(ymmetric)
• $\gamma^5$ is Hermitean A(ntisymmetric) $\leadsto$ check this from the above!

Thus $$ \bar{\epsilon}\eta= i\epsilon^T \gamma^0 \eta =-i\eta^T \gamma^{0~~T} \epsilon =i\eta^T \gamma^{0 }\epsilon = \bar{\eta} \epsilon , $$ So now you got your baseline. Make sure you appreciate every step.

The rest follow trivially from the above properties of the γ matrices and the above properties, so supplanting $\gamma^0$, you now have $$(\gamma^0 \gamma^5)^T= \gamma^5 \gamma^0= -\gamma^0 \gamma^5,$$ just like for the scalar, so $\bar{\epsilon}\gamma_{5}\eta=\bar{\eta}\gamma_{5}\epsilon$. (Anti-)likewise, $$ (\gamma^0 \gamma^\mu)^T= \gamma^0 \gamma^\mu, $$ where you can marvel at how the time-like and space like parts synchronize to yield a uniform answer, hence $\bar{\epsilon}\gamma^{\mu}\eta=-\bar{\eta}\gamma^{\mu}\epsilon$. Finally, for the axial, $\gamma^5$ flips the sign to yield $\bar{\epsilon}\gamma^{\mu}\gamma_{5}\eta=\bar{\eta}\gamma^{\mu}\gamma_{5}\epsilon$ .