Hi,

I'm not sure I completely understood what you mean by defining a new TM $M_j$…

The idea is this:

You have in infinite list of TMs $M_1, M_2, ...$.

In question 1:

Let's build a deciding TM $M_A$ for A - the TM will simply return "True".

Why is this true? there are infinite many TMs that accepts $\Sigma^*$.

Therefore, no matter which $M_i$ is received as input for $M_A$, there will always be $j > i$ such that $M_j = \Sigma^*$.

In question 2:

Let's assume for simplicity that $L(M_1) = \Sigma^*$. Then we're done - for every $M_i$ there exists $j < i$ (specifically, $j = 1$) s.t. $L(M_j) = \Sigma^*$.

Of course, this doesn't have to be true. It might be the case that $L(M_1) \neq \Sigma^*$. Same goes for $M_2, M_3,...$.

However, at some point, call it $j_0, L(M_{j_0}) = \Sigma^*$.

Therefore, for all $i > j_0$, there exists $j < i$ (specifically, $j = j_0$) s.t. $L(M_j) = \Sigma^*$.

For the first $j_0-1$ TMs there is no TM as requested (since we assumed $M_{j_0}$ is the first TM that accepts $\Sigma^*$), but in any case this is only a finite set - those machines belong to $\overline{B}$.