Covariant derivatives of the frame field, are they the same?

As far as I known the anti-symmetric part of (2) equals to (1).

You first statement looks like the definition of the torsion form $$ D{bf e}^{*a}equiv d{bf e}^{*a}+ {omega^a}_b wedge {bf e}^{*b}= T^a. $$ The second is a confused aspect of the ordinary covariant derivative and has nothing to do torsion or metricity.

The $e_a^mu$ and so on are just an array of numbers. Their covariant derivative is just the ordinary partial derivative. The expression you have is therefore not the covaraint derivative of $e^mu_a$. It is the geometric basis vectors ${bf e}_a$ or ${boldsymbol partial}_mu$ for the tangent space, or ${bf e}^{*a}$, or $dx^mu$ for the cotangent space that need the connection forms ${omega^b}_{amu}dx^mu$ or ${Gamma^lambda}_{numu}dx^mu$.

In particular the "vierbein postulate" is not a postulate. It is an identy that always holds. It is in fact an appallingly bad and confusing name for the formula for converting the covariant derivative $nabla_Xequiv X^mu nabla_mu$ from its definition in terms of a frame-field $$ nabla_X {bf e}_a= {bf e}_b{omega^b}_{amu}X^mu $$ to its definition in terms of the coordinate basis
$$ nabla_X {boldsymbol partial }_nu= {boldsymbol partial }_lambda {Gamma^lambda}_{numu} X^mu. $$ In each case $nabla_X$ is the same object. So by expanding
$$ {bf e}_a = e^nu_a {boldsymbol partial}_nu $$ and using the derivation property (Leibnitz rule) of $nabla_X$, we evaluate $$ nabla_X {bf e}_a = e_b^nu {omega^b}_{amu}X^mu{boldsymbol partial}_nu $$ in the equivalent form $$ nabla_X {bf e}_a = X^mu (partial_ mu e^nu_a){boldsymbol partial}_nu + X^mu e^nu_a (nabla_mu {boldsymbol partial}_nu) =X^mu (partial_mu e_a^nu+ e^lambda_a {Gamma^nu}_{lambda mu}){boldsymbol partial}_nu. $$ By comparing coefficients of $X^mu {boldsymbol partial}_nu$, we read off that
$$ partial_mu e_a^nu+ e^lambda_a {Gamma^nu}_{lambda mu}- e_b^lambda {omega^b}_{amu}=0. $$ Your equation (2) is just an example of this manipulation for the co-frames ${bf e}^{*a}= e^{*a}_mu dx^mu$.