No, your $L$ (which is sometimes called $\text{REC}$) is $\Sigma_3^0$-complete. However, this proof is somewhat technical. If you just want to show that it and its complement is not c.e., there is a fairly easy proof:
It well known that $K = \{e : \Phi_e(e) \downarrow\}$ is c.e. and but not computable. ($K$ is a variant of the Halting Problem.) It is easy to show that if a set and its complement is c.e., then it must be computable. Hence the complement of $K$, denoted $\bar{K}$, is not c.e.
Define
$\Psi(e,x) = \begin{cases}
0 & \quad \Phi_{e,x}(e)\uparrow \\
0 & \quad \Phi_{e,x}(e) \downarrow \wedge (\exists s)\Phi_{x,s}(x)\downarrow \\
\uparrow & \quad \Phi_{e,x}(e) \downarrow \wedge \Phi_{x}(x)\uparrow
\end{cases}$
$\Psi$ is clearly a partial computable function. By the s-m-n theorem, there exists a computable $f$ such that $\Psi(e,x) = \Phi_{f(e)}(x)$. Thus if $e \in \bar{K}$, then $\Psi(e,x) = \Phi_{f(e)}(x) = 0$ for all $x$ so $W_{f(e)} = \omega$. $\omega$ is computable. So $f(e) \in L$. If $x \in K$, then there is a least $M$ such that $\Phi_{e,M}(e) \downarrow$. So then for all $x > M$, $\Psi(e,x)\downarrow = \Phi_{f(e)}(x)\downarrow$ if and only if $x \in K$. So $W_{f(e)}$ is equal to $K$ except for a finite difference. Since $K$ is not computable, $W_{f(e)}$ is also not computable. Hence $f(e) \notin L$. Thus it has been shown $f$ is the reduction of $\bar{K} \leq_m L$. Since $\bar{K}$ is not c.e., $L$ can not be c.e.
To show that the complement $\bar{L}$ is not c.e. Use
$$\Psi(e,x) = \begin{cases}
0 & \quad \Phi_{e,x}(e) \uparrow \wedge (\exists s)\Phi_{x,s}(x)\downarrow \\
\uparrow & \quad \Phi_{e,x}(e)\uparrow \wedge \Phi_x(x)\uparrow \\
0 & \quad \Phi_{e,x}(e)\downarrow
\end{cases}$$
By a similar argument as above, $\bar{K} \leq_m \bar{L}$. Hence $\bar{L}$ can not be c.e. either.
It has been shown that $L$ and $\bar{L}$ is not c.e.
The basic idea is: The set $K = \{\langle M \rangle : M(\langle M \rangle) \text{ halt }\}$ is recognizable but not decidable. Hence its complement $\bar{K} = \{\langle M \rangle : M(\langle M \rangle) \text{ does not halt}\}$ is not recognizable.
Let $M_x$ denote the Turing machine with code $x$. You want to construct a Turing machine $U$ that reduces $\bar{K}$ to $L$. Given $M$ define $T_M$ to the be Turing machine such that:
$T_M(x)$ halts if $M(\langle M \rangle)$ does not halt by stage $x$
$T_m(x)$ halts if $M(\langle M \rangle)$ halts by stage $x$ and $M_x(x)$ eventually halts
$T_M$ does not halt otherwise
Let $U(\langle M \rangle) = \langle T_M \rangle$
Then if $\langle M \rangle \in \bar{K}$, $T_M$ halts on all input. Hence $L(T_M)$ is decidable. So $U(M) = \langle T_M \rangle \in L$ If $\langle M \rangle \notin \bar{K}$, then $L(T_M)$ agrees with $K$ except on a finite set. Since $K$ is not decidable, $L(T_M)$ is not decidable. So $U(M) = \langle T_M \rangle \notin L$. This give the reduction of $\bar{K} \leq L$. Since $\bar{K}$ is not recognizable, $L$ is not either.
Using a similar idea, you can reduce $\bar{K} \leq \bar{L}$, where $\bar{L}$ is the complement of $L$.
Suppose T is decided by R.
Construct the following machine H.
$H(<A>, x):$
1) Create the following machine $< M >:$
$\quad\quad M(w):$
$\quad\quad\quad A(x)$
$\quad\quad\quad accept$
2) if $R(<M, w>)$ accepts then
$\quad\quad accept$ (since A(x) halts)
$\;\;\;$else
$\quad\quad reject$ (since A(x) loops)
Note that $<M>$ is constructed to accept all w if and only if A(x) halts.
In that case, $<M>$ certainly accepts w and reverse(w).
So $<M, w>$ is in T if and only if A(x) halts.
If R is a decider, then H would be a decider for the Halting problem.
So, T must be undecidable.
Now, construct the recognizer for T:
$R(<M, w>):$
$\quad$ if M(w) accepts and M(reverse(w)) accepts then
$\quad\quad$ accept
$\quad$ else
$\quad\quad$ reject
Note that if either M(w) or M(reverse(w)) rejects or loops then R will reject or loop.
So, R is a recognizer for T.
Best Answer
For 1., you're right that it follows directly from Rice theorem. Reduction to REGULAR seems hard to do: you would have to build a machine $M'$ from a machine $M$, such that $L(M')$ is finite if and only if $L(M)$ is regular.
For 2., any machine can be turned into a machine with 1 accepting state, and then deciding whether this state is reachable is the same as deciding whether $L(M)\neq\emptyset$. So it shows that USELESS' is undecidable. Then, you can show that USELESS is also undecidable in the following way:
Suppose you have an algorithm for USELESS, then you can always try removing 1 state, until the machine is not in USELESS. If it does not work try 2 states, and so on... This way you will find the set of useless states. If the accepting state is in them, then the machine does not accept anything, otherwise it does. This shows that USELESS is also undecidable.