Proving Infinite Turing Machine Language (with finite subset) is Recursively Enumerable
Regarding your first question, a reduction from the halting problem only shows that a problem is hard, not that it's easy. If the halting problem reduces to some problem $P$, then you know that $P$ isn't co-computably enumerable. So if you manage to reduce the halting problem to $L$, then this shows that $L$ is not co-computably enumerable. Perhaps that can inform your guess on the answer.

Regarding your second question, if you show that $L$ is c.e. but not computable, then indeed it follows directly that $L$ is not co-c.e. This is because a language that is both c.e. and co-c.e. is computable.

Let me give you a hint. Every Turing machine can be modified so that it accepts only strings beginning with 0: the new machine accepts $0x$ iff the old machine accepts $x$. Such a Turing machine belongs to $L$ iff it accepts infinitely many strings. Similarly, a machine which only accepts strings beginning with 1 belongs to $L$ iff it accepts finitely many strings.