You're not getting your facts right at all.
How do we know from this $\langle W \rangle = \int_{-\infty}^{\infty} \bar{\Psi}\left(-\frac{\hbar^2}{2m} \frac{d^2}{dx^2} + W_p \right) \Psi dx$ or this $\hat{H} = -\frac{\hbar^2}{2m}\frac{d^2}{dx^2} + W_p$ that we have an eigenfunctiuion and eigenvalue.
Answer: we don't.
All I know about operator $\bar{H}$ so far is this equation where $\langle W \rangle$ is an energy expected value:
\begin{align}
\langle W \rangle = \int_{-\infty}^{\infty} \bar{\Psi}\left(-\frac{\hbar^2}{2m} \frac{d^2}{dx^2} + W_p \right) \Psi dx
\end{align}
No, you don't.
Here's the mathematical side of what an eigenfunction and eigenvalue is:
Given a linear transformation $T : V \to V$, where $V$ is an infinite dimensional Hilbert or Banach space, then a scalar $\lambda$ is an eigenvalue if and only if there is some non-zero vector $v$ such that $T(v) = \lambda v$.
Here's the physics side (i.e. QM):
We postulate that the state of a system is described by some abstract vector (called a ket) $|\Psi\rangle$ that belongs to some abstract Hilbert space $\mathcal{H}$.
Next we postulate that this state evolves in time by some Hermitian operator $H$, which we call the Hamiltonian, via the Schrodinger equation. What is $H$? you guess and compare to experimental results (that's what physics is anyway).
Next we postulate for any measurable quantity, there exists some Hermitian operator $O$, and we further postulate that the average of many measurements of $O$ is given by $ \langle O \rangle = \langle \Psi | O | \Psi \rangle$.
Connection to wavefunctions: we pick the Hilbert space $L^2(\mathbb{R}^3)$ to work in, so $\Psi(x) = \langle x | \Psi \rangle$, and $\langle O \rangle = \int_{-\infty}^{\infty} \Psi^*(x) O(x) \Psi(x) dx$.
Ok, that's the end. The form of $H$ doesn't follow from the energy expected value.
Wait! I haven't even talked about eigenvalues and eigenfunctions. This is a useless post!
Answer: well you don't have to. But it is useful to find the eigenvalues and eigenfunctions of $H$, because the eigenfunctions of $H$ form a basis of the Hilbert space, and certain expressions become diagonal/more easily manipulated when we do whatever calculations we want to do.
So to find the eigenvalues of $H$, we simply solve the eigenvalue equation as stated above:
Solve
\begin{align}
H | \Psi_n \rangle = E_n | \Psi_n \rangle.
\end{align}
This is in the form $T(v) = \lambda v$.
So as Alfred Centauri says, we simply want to find the eigenfunctions of $H$. A more subtle question would be, how do we know they exist? The answer lies in spectral theory and Sturm-Liouville theory but nevermind for now, as physicists we assume they always exist.
So your additional question:
$\hat{a} \psi$ is an eigenfunction of operator$\hat{H}$ with
eigenvalue $(W-\hbar \omega)$.
Well.... that just follows straightaway. You said you already proved that $H a^\dagger \psi = (W - \hbar \omega) a^\dagger \psi$. So here $T$ = $H$, $a^\dagger \psi = v$, and $\lambda = (W - \hbar \omega)$. which is an eigenvalue equation $T(v) = \lambda v$. Thus, $a^\dagger \psi$ is an eigenfunction of $H$ with eigenvalue $(W-\hbar \omega)$.
For the specific case of the particle-in-a-box problem, the momentum operator is much trickier to handle than what you're allowing for, but that's not the issue at the heart of your current confusion.
If you know that two operators $A$ and $B$ commute, i.e. $[A,B]=0$, then there's two common ways to understand the consequences, one of which is right and one of which is wrong:
You are guaranteed the existence of at least one shared eigenbasis of both operators; but
You are not guaranteed that all eigenbases of $A$ will be eigenbases of $B$, or vice versa.
(For other occurrences of the misconception that the second point does happen, see e.g. this answer or this one.)
Thus, the 'contradiction' you observe is also present for the case of the free particle, without the box: the wavefunction $\psi(x)=\cos(kx)$ is an eigenfunction of $\hat H=\frac{1}{2m}\hat P^2$, but it is not an eigenfunction of $\hat P$. That is, the fact that the hamiltonian's eigenfunctions over a compact interval are not momentum eigenfunctions is not surprising and it is not specific to that configuration.
But, that said, you can still ask something like
well, OK, so some $\hat H$ eigenfunctions in the particle-in-a-box problem are not $\hat P$ eigenfunctions and that's not a problem, but $[\hat H,\hat P]=0$ is still true, so shouldn't I be guaranteed a shared eigenbasis, even if it isn't the one I started with?
and it's a reasonable question. Here what happens is that the subtleties with the compact-interval momentum operator kick in, at a much deeper level than just the commutation: the result, in full, reads
If two self-adjoint operators $A$ and $B$ commute, then there exists at least one shared eigenbasis,
and it breaks because $\hat P$-in-a-box is not a self-adjoint operator: it is symmetric, but it has domain problems that prevent it from being self-adjoint. The consequences of this are as deep as they get: there simply isn't a momentum eigenbasis in this Hilbert space. You can extend the momentum operator to make it self-adjoint; this extension is not unique, but there is a reasonable choice (setting $\alpha=0$ in V. Moretti's answer) that comes close to making $\hat H$ the square of the extended $\hat P_\alpha$, but ultimately this can't work, as they have different domains. (More specifically, the domain of $\hat H$ is contained in the domain of $\hat P_\alpha$, but the eigenfunctions of $\hat P_\alpha$ do not fall into that subspace.)
Best Answer
This is a good question (or at least I think so because I struggled with the same question when I first studied quantum mechanics).
Notational Clarity
First of all, let's establish some notation. I will call the vector-space $\mathbb{V}$ and I will assume that it is finite-dimensional. Nothing of relevance hinges upon its finity and it'll save me some time. I will denote vectors in the vector-space as $\vert v\rangle$. Now, there are three different $0$s that you are playing with.
Onto Your Questions
Mathematically, there is nothing too confusing here. As already pointed out in the comments, $0\vert v\rangle=\vert\Phi\rangle,\forall\vert v\rangle\in\mathbb{V}$. And thus, clearly, if $A\vert\psi\rangle=\vert \Phi\rangle$ the eigenvalue equation is satisfied by $\vert\psi\rangle$ with the eigenvalue being $0$.
Look at it this way if it helps: The eigenvalue equation says that if the two quantities $A\vert\psi\rangle$ and $e\vert\psi\rangle$ are equal then $\vert\psi\rangle$ is an eigenstate of $A$ with the eigenvalue $e$. OK, so, let's check if this criterion is satisfied for an eigenstate with zero eigenvalue: The first quantity is $A\vert\psi\rangle=\vert\Phi\rangle$ and the second quantity is $0\vert\psi\rangle = \vert \Phi\rangle$. Voila! The criterion is obviously satisfied.
Now, it is completely irrelevant that $0\vert \lambda\rangle=\vert \Phi\rangle$ for $\vert\lambda\rangle\neq\vert\psi\rangle$. The eigenvalue equation does not say that there cannot exist a $\vert\lambda\rangle\in\mathbb{V}$ such that $A\vert\psi\rangle=e\vert\lambda\rangle$ for $\vert\psi\rangle$ to be an eigenstate of $A$ with the eigenvalue $e$.
No, you're misunderstanding one of the two things here:
I am not sure how this tells you what you think it tells you but you are making a mistake of confusing the scalar and the vector zeroes. In particular, the correct version is $\hat{H}\vert 0\rangle = 0 \vert 0\rangle = \vert\Phi\rangle $ and $a\vert 0\rangle = \vert \Phi\rangle$. In your notation, my $\vert\Phi\rangle$ is $\vec{0}$. Hope this helps.