$(x \in K)$ or $(x \in W_e)$ does not count as a bounded quantifier in Computability Theory where bounded means bounded by a number. Note this is different than in the first order theory of Set theory.
For all of these, my Halting Problem or Jump $K$ is defined as $K = \{e : \Phi_{e}(e) \downarrow\}$. The notation $\Phi_{e,s}(x)$ means run the $e^\text{th}$ Turing Program for $s$ steps on input $x$. The important part is that this is computable.
- On the surface, $A_1$ is $\Pi_1^0$.
$A_1 = \{e : (\forall n)(\forall s)(\Phi_{e,s}^\emptyset(2n)\uparrow)\}$
This is $\Pi_1^0$. In fact, it well known that $K$ is the $\Sigma_1^0$ 1-complete (complete via 1-reductions). Therefore, the complement of $K$ is $\Pi_1^0$ 1-complete. The claim is that $A_1$ is also $\Pi_1^0$ 1-complete. Define the function $f$ as follows :
$\Phi_{f(e)}(x) = \begin{cases} 1 & \quad x = 0 \wedge (\Phi_e(e)\downarrow) \\ \uparrow & \quad \text{otherwise} \end{cases}$
By some theorem (maybe the s-m-n theorem), the function $f$ exists and is injective and used to prove the 1-reduction $\bar{K} \leq_1 A_1$. That is, if $e \in \bar{K}$, then $W_{f(e)} = \emptyset$. Thus $f(e) \in A_1$. If $e \in K$, then $W_{f(e)} = \{0\}$, then $f(e) \notin A_1$. Thus $\bar{K} \equiv_1 A_1$.
For the second one, one can write
$A_2 = \{e : (\exists x)(\exists s)(\Phi_{x,s}(x)\downarrow)\}$
This is $\Sigma_1^0$. This set is also $\Sigma_1^0$-complete. Define the function $f$ as follows :
$\Phi_{f(e)}(x) = \begin{cases} 1 & \quad x = e \\ \uparrow & \quad \text{otherwise} \end{cases}$
Note that $W_{f(e)} = \{e\}$. Thus, this function witnesses $K \leq_1 A_2$.
The third one as far as I can tell is $\Pi_2^0$
$A_3 = \{e : (\forall n)((\exists s)(\Phi_{n,s}(n)\downarrow) \Rightarrow (\exists t)(\Phi_{e,t}(n)\downarrow))\}$
I believe $A_3$ is $\Pi_2^0$ 1-complete. $A_4$ or $Inf$ is $\Pi_2^0$ complete (as I will say again below). I will show $A_3$ is $\Pi_2^0$ complete by reducing $Inf$ to it. Define the function $f$
$\Phi_{f(e)}(x) = \begin{cases} 1 & \quad (\exists s)(\exists t)(s > x \wedge \Phi_{e,t}(s)\downarrow \wedge \Phi_{e,s}(e)\downarrow \\ \uparrow & \quad \text{otherwise} \end{cases}$
I believe this function witness that $Inf \leq_1 A_3$. If $e \in Inf$, then $W_{f(e)}$ contains $K$ since if $x \in K$ there exists a $s > x$ such that $s \in W_e$ and $\Phi_{e,s}(e)\downarrow$. Thus $x \in W_{f(e)}$. If $e \notin Inf$, then $W_e$ is finite. Let $j$ be the largest element in $W_e$. Since $K$ is infinite, it is clear that $W_{f(e)}$ does not contain $K$ since $W_{f(e)}$ is bounded by $j$.
The fourth one is well known. It is often called Inf
$A_4 = Inf = \{e : (\forall n)(\exists x)(\exists t)(x > n \wedge \Phi_{e,t}(x)\downarrow)\}$
$Inf$ is $\Pi_2^0$ complete. Look in Soare's book for a proof.
If my proofs are correct, all you sets above can certainly be defined by a $\Pi_1^0$, $\Sigma_1^0$, $\Pi_2^0$ and $\Pi_2^0$ formula, respectively, but in some sense you can even prove the lowest they can be on the hierarchy are these places.