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 $L$.
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.
Best Answer
It suffices to focus on the case $\Sigma = \{0,1\}$. We argue that $ALL_{\textsf{minCFG}} \leq_T MIN_{\textsf{CFG}}$. That is, we can determine whether a minimal CFG can generate all possible strings, by using the oracle for $MIN_{\textsf{CFG}}$.
Let $G$ denote the minimal CFG we are interested. Since $G$ is minimal, there exists a length $p$ such that, for each rule $R$ in $G$ there exists a string $w_R$ with $|w_R|<p$ that can be generated by $G$, but not with $R$ deleted. Note here that $p$ is computable in a brute force manner. Construct $G^+_{a_1a_2\cdots a_p}$ where $a_i \in \Sigma$, by adding following rules in $G$: $$ S \to a_1 a_2 \cdots a_pT \\ T \to T0 \ | \ T1 \ | \ \epsilon $$ First, consider what will happen if $G^+_{a_1a_2\cdots a_p}$ is not minimal, thus, there exists a rule $R$ in $G^+$ being dispensable. By our construction it is clear that each rule $R$ in $G$ is indispensable.
Therefore, if $G^+_{a_1a_2\cdots a_p}$ is not minimal, we can determine whether $a_1 a_2 \cdots a_p \Sigma^* \subseteq L(G)$. What if it is minimal? In this case we will immediately know $L(G) \neq \Sigma^*$. So we can suppose that all possible $G^+_{a_1a_2\cdots a_p}$ is not minimal. Then by checking whether $a_1 a_2 \cdots a_p \Sigma^* \subseteq L(G)$ one by one and checking those strings of less-than-$p$ length, we can still decide whether $L(G) = \Sigma^*$.
Update. (Oct 12, 2020) After a short discussion with @xdavidliu I realized that the argument above can directly prove what we want, without a justification of uncomputability of $ALL_{\textsf{minCFG}}$. Recall that the reduction we have established achieves the following: Given a minimal CFG $G$, determine whether $L(G) = \Sigma^*$. Now if we want to decide whether $L(G_0) = \Sigma^*$ for an arbitrary CFG $G_0$, just simply feed into the (reduction) algorithm all minimal sub-CFG (test each sub-CFG by the $MIN_{\mathsf{CFG}}$ oracle to determine whether it is minimal) of $G_0$ and see if there is one that generates all strings.
Since this problem appears in Chapter 5, in which Turing reduction has not introduced. I personally believe that there is an elegant proof using only mapping reduction.
Update. @Sylvain gives an elegant proof on https://cstheory.stackexchange.com/a/39454/46760, by showing $$ PCP \leq_T FullPCP \leq_m MIN_{\textsf{CFG}}$$ in which I personally thought the key is that context-free language is closed under $h^{-1}$ he defined.