Whenever you approach one of these problems, the first thing to do is what Rogers calls a Tarski-Kuratowski computation: find the level of the arithmetical hierarchy where the set in question lives. In this case, we have that $e \in \text{Tot}$ if for every $n$ there exists an $s$ with $\phi_{e,s}(n)\downarrow$ (that is, $\phi_e(n)$ halts in $s$ steps). Therefore Tot is $\Pi^0_2$ and its complement is $\Sigma^0_2$.
So how can we diagonalize against Tot to show it is not computably enumerable? Well, we can use the leading $\forall$ quantifier in its definition, and the fact that the halting set $K$ is already $\Sigma^0_1$, to show that if Tot was $\Sigma^0_1$ then the halting set would be computable. Recall that, in general, a set is computable if both it and its complement are $\Sigma^0_1$, which is equivalent to the set and its complement both being $\Pi^0_1$.
Let $e$ be given. We want to tell whether $e \in K$, that is, whether $\phi_e(0)$ halts. Let $g$ be such that $g(s) = 0$ if $\phi_{e,s}(0)\uparrow$ and $g(s)$ diverges if $\phi_{e,s}(0)\downarrow$. Notice that $g$ is a partial computable function, and $g$ is total if and only if $\phi_e(0)\uparrow$, and we can compute an index $\gamma_e$ for $g$ uniformly from an index $e$. Therefore, we have
$$
e \in K \leftrightarrow (\exists s)[\phi_{e,s}(0) \downarrow]\\
e \not \in K \leftrightarrow \gamma_e \in \text{Tot}
$$
So, if Tot was computably enumerable, the two facts there the would show that $K$ is computable, which is impossible. The method here was to leverage the $\forall$ quantifier in Tot; if we had a way to decide membership in Tot, that gives us a "free pass" to decide that universal quantifier. The definition of $\gamma_e$ leverages the universal quantifier to tell whether $\phi_e(0)$ does not halt.
How can we show that Tot is not $\Pi^0_1$? Well, we can use the inner $\exists$ quantifier in the definition of Tot, and the fact that the complement $\bar K$ of the halting problem is not computable. In the first part we used Tot to tell whether $\phi_e(0)\uparrow$. Now we will use Tot to tell whether $\phi_e(0)\downarrow$. Given $e$, let $h(x) = \mu s [\phi_{e,s}(0) \downarrow]$. Then $h$ is total if and only if $\phi_e(0)\downarrow$, which is equivalent to $e \not \in \bar K$. Also, we can compute an index $\eta(e)$ for $h$ uniformly from $e$. Thus:
$$
e \in \bar K \leftrightarrow (\forall s)[\phi_{e,s}(0)\uparrow]\\
e \not \in \bar K \leftrightarrow \eta(e) \in \text{Tot}
$$
Therefore, if $\text{Tot}$ was $\Pi^0_1$ then $\bar K$ would be computable, which is impossible.
To try to determine whether a language is finite, we have to have some information about the language. So, to make the problem precise, it's necessary to specify how that information will be provided. Perhaps the most typical way to pose this sort of question is the following:
Given a program $e$ for a Turing machine, is the language accepted by machine $e$ finite?
So we are asking whether that problem can be solved by a computable function:
Is there a computable function $f$ such that, given a program $e$ for a Turing machine, $f(e) = 1$ if the language accepted by machine $e$ finite, and $f(e) = 0$ otherwise?
The answer to that is "no". There is a general result, Rice's theorem, which applies. It shows that, when we ask questions like this one about the language accepted by a Turing machine (i.e. not questions about the machine itself), no nontrivial property of languages can be decided by a computable function that takes as input the machine $e$.
The general proof method is by reducing the halting problem to the question above. If we want to tell whether a particular Turing machine $e$ halts when started on an empty tape, we can make another Turing machine $e'$ which accepts a finite language if and only if $e$ halts. The function taking $e$ to $e'$ is computable. So if $f$ above was computable, we could solve the halting problem by taking input $e$ and asking whether $f(e') = 1$. That would let us solve the halting problem with a computable function, which is impossible. We conclude that $f$ is not computable, because the map taking $e$ to $e'$ is.
Best Answer
Say your $L$ was recursive. Then it's complement would also be recursive, i.e. the problem of whether some $M_w$ diverges for all inputs would be decidable.
Now, say you have a turing machine $M'$ and some input $x'$. You can then find some turing machine $M$ which first writes $x'$ to the tape (overwriting the previous input), and then continues as $M'$ would. $M$ diverges for all inputs exactly if $M'$ diverges for input $x'$. Thus, if you can decide whether or not a arbitrary turing machine $M$ diverges for all inputs, you can also decide whether an arbitrary machine terminates for some arbitrary input. Which you know that you cannot, since this is the halting problem. $L$ is thus not recursive.
$L$ is recursively enumerable, though, because you can simply interate over all triples $(M_w,x,n)$ (diagonally!) and check whether $M_x(x)$ converges after $n$ steps. If $w \in L$, you'll eventually reach the input $x$ for which it converges and some $n$ which is at least as large as the number of steps it takes for $M_w(x)$ to converge, thus discovering that $w \in L$.