Using Generalized Rice's Theorem to Prove Decidability
I'm going to ignore the second half of your post, and answer the question that appears at the first half. If this is not what you meant - post another question or edit as necessary.

Given a machine $M$ we define $L(M)$ to be the language recognized by that machine. In other words, $L(M)$ is the set of all the words $w\in \Sigma^*$ such that $M$ accepts $w$.

Thus, by its definition $L(M)$ is at least semi-decidable (because $M$ is accepting it!!)

To clarify whether $L(M)$ is decidable or semi-decidable(but not fully decidable), we need more information on $M$.

For instance, it can be that $M$ is the trivial machine that immediately go to the accepting state. In that case $L(M)=\Sigma^*$ (all the possible words), and $L(M)$ is decidable.

On the other hand it can be that $M$ does the following:   

1. it ignores the first symbol of the input
2. it runs HP on the rest of the input


where HP is the machine that accepts the Halting Problem language. It is not too difficult to see that in this case $L(M)$ cannot be fully decidable (or otherwise the halting problem will be decidable), but, as said above, it is semi-decidable by definition (that is, it is in $RE\setminus R$).



Now, if you don't care about a specific machine $M$, but instead about all the machines that accept infinitely many words that begin with 1 and infinitely many words that begin with 2, the language that contain all such machines cannot be decidable due to Rice's theorem.

Just to be clear on the statement: let $S \subset RE$ be all the languages that contain infinitely many words that start with 1 as well as infinitely words that start with 2, then

$$L_S = \{ \langle M \rangle \mid L(M) \in S \}$$ cannot be decidable, $L_S\notin R$. Note that $L_S$ contains all the (encodings of) machines that satisfy the property $S$  defined in your question.