My apologies about the lack of type-setting software!

The answer to part (a) of your question is that the normal closure of a^k1.t.a^k2.t.a^k3 is always equal to the normal closure of a and t^2.
This is because the quotient of G obtained by adding a relation of the type written in part (a) always gives a=1!

Let me continue to work from now on in the quotient (and by abuse of notation with the same letters a and t.) The added relation can be written in the form t.a^m.t.a^n=1, so t.a^m=a^-n.t^-1. 
Then b=t.a.t^-1=(t.a^m)a(t.a^m)^-1, so also b=(a^-n.t^-1).a.(t.a^n) and hence t^-1.a.t=a^n.b.a^-n.

It follows that not only does one have b.a.b^-1=a^2, but (after conjugating all letters in the last written relation through by a^-n) we can hence deduce (t^-1.a.t)a(t^-1.a.t)^-1=a^2. Now conjugating (every letter of) this last relation through by t we obtain a.b.a^-1=b^2. 

It is easy to see that b.a.b^-1=a^2 and a.b.a^-1=b^2 imply a=1.