Indirect Left recursion
Yes, both your answers are correct. Note that the language under consideration is regular, see wikipedia.

The first grammar is equivalent to this regular expression:
$$r_1 = a(ba)^* \mid (ba)^+.$$

The second grammar is equivalent to this regular expression:
$$r_2 = (ab)^+a \mid b(ab)^*a \mid a.$$

It's easy to prove those regexes generate equal languages $L(r_1) = L(r_2)$.