First of all you should recall that Schroedinger equation is an Eigenvalue equation. If you are unfamiliar with eigenvalue equations you should consult any math book or course as soon as possible.
Answer 1 (my apologies, I will use my own notation, as this is mainly copy-paste from my old notes):
First define constants
\begin{equation}
x_0 = \sqrt{\frac{\hbar}{m\omega}} ,
\end{equation}
\begin{equation}
p_0 = \frac{\hbar}{x_0} = \sqrt{\hbar m \omega} ,
\end{equation}
and dimensionless operators
\begin{equation}
\hat{X} = \frac{1}{x_0} \hat{x} ,
\end{equation}
and
\begin{equation}
\hat{P} = \frac{1}{p_0} \hat{p} .
\end{equation}
Their commutation relation then is
\begin{equation}
\left[ \hat{X} , \hat{P} \right] = \left[ \frac{1}{x_0}\hat{x} , \frac{1}{p_0}\hat{p} \right] = \frac{1}{x_0 p_0} \left(\hat{x} \hat{p} - \hat{p} \hat{x} \right) = \frac{1}{x_0 p_0} \left[ \hat{x} , \hat{p} \right] = \frac{i\hbar}{x_0 p_0} = i ,
\end{equation}
as
\begin{equation}
x_0 p_0 = \sqrt{\frac{\hbar}{m\omega}} \sqrt{\hbar m\omega} = \hbar .
\end{equation}
Now write Hamiltonian in terms of $\hat{X}$ and $\hat{P}$. Start with
\begin{equation}
\hat{H} = \frac{p_0 ^2}{2m} \hat{P} ^2 + \frac{1}{2} m\omega^2 x_0 ^2 \hat{X}^2 .
\end{equation}
Notice that
\begin{equation}
p_0 ^2 = \hbar m \omega
\end{equation}
and
\begin{equation}
x_0 ^2 = \frac{\hbar}{m \omega} ,
\end{equation}
hence
\begin{equation}
\hat{H} = \frac{\hbar \omega}{2} \hat{P}^2 + \frac{\hbar \omega}{2} \hat{X}^2 = \frac{\hbar \omega}{2} \left(\hat{X}^2 + \hat{P}^2 \right).
\end{equation}
Up to the commutation relation we can write
\begin{equation}
\left(X^2 + P^2 \right) = \left(X - iP \right) \left(X + iP \right) .
\end{equation}
On the other hand, for operators this is not quite allowed, as
\begin{align}
\left(\hat{X} - i\hat{P} \right) \left(\hat{X} + i\hat{P} \right) &= \hat{X}^2 + i\hat{X}\hat{P} - i\hat{P}\hat{X} + \hat{P}^2 \nonumber \\ &= \hat{X}^2 +i \left(\hat{X}\hat{P} - \hat{P}\hat{X}\right) + \hat{P}^2 \\ &= \hat{X}^2 + i \left[ \hat{X} , \hat{P} \right] + \hat{P}^2 = \hat{X}^2 + \hat{P}^2 - 1 \nonumber ,
\end{align}
so one has
\begin{equation}
\left(\hat{X}^2 + \hat{P}^2 \right) = \left(\hat{X} - i\hat{P} \right) \left(\hat{X} + i\hat{P} \right) + 1 .
\end{equation}
Now we can define
\begin{equation}
\hat{a} = \frac{1}{\sqrt{2}} \left(\hat{X} + i\hat{P} \right) ,
\end{equation}
and
\begin{equation}
\hat{a}^\dagger = \frac{1}{\sqrt{2}} \left(\hat{X} - i\hat{P} \right),
\end{equation}
calling this creation operator and $\hat{a}$ - annihilation operator. Notice that we can now express Hamiltonian in terms of creation and annihilation operators:
\begin{equation}
\hat{H} = \frac{\hbar\omega}{2} \left(\sqrt{2} \hat{a} ^\dagger \sqrt{2} \hat{a} + 1 \right) = \hbar \omega \left(\hat{a} ^\dagger \hat{a} + \frac{1}{2} \right) .
\end{equation}
But we can also define the number operator, $\hat{N} = \hat{a} ^\dagger \hat{a}$, so finally get
\begin{equation}
\hat{H} = \hbar \omega \left(\hat{N} + \frac{1}{2} \right) .
\end{equation}
Now go aside a bit and consider creation and annihilation operators. By definition,
\begin{equation}
\hat{a}^\dagger \left| n \right\rangle = \sqrt{n + 1} \left| n + 1 \right\rangle ,
\end{equation}
\begin{equation}
\hat{a} \left| n \right\rangle = \sqrt{n} \left| n - 1 \right\rangle ,
\end{equation}
where $\left| n \right\rangle$ is the eigenstate of creation and annihilation operators, as well as of the Hamiltonian (due to the fact that they commute - homework to prove).
Now
\begin{equation}
\hat{a}^\dagger\hat{a} \left| n \right\rangle = \hat{a}^\dagger \sqrt{n} \left| n - 1 \right\rangle = \sqrt{n} \sqrt{n} \left| n \right\rangle = n \left| n \right\rangle ,
\end{equation}
so conclude that the eigenvalue of a number operator, $\hat{N}$, is just $n$, so if we now apply Hamiltonian in the Schroedinger equation, get
\begin{equation}
\hat{H} \psi = E \psi ,
\end{equation}
\begin{equation}
E_n = \hbar \omega \left(n + \frac{1}{2} \right) ,
\end{equation}
which is exactly the result you were looking for.
Answer 2:
First of all you should remember that the general aim of solving an eigenvalue problem is to find a set of eigenvectors, but not a single eigenvector. In your case, equation should be modified to
\begin{equation}
\frac{d^2 \psi_n}{dx^2} + \left[(2n + 1) - \varepsilon^2\right]\psi_n = 0 ,
\end{equation}
where $\psi_n$ are eigenvectors (eigenfunctions) that correspond to eigenvalues $E_n$. Try to think a little bit and explain physical meaning of having many energy eigenvalues in quantum mechanics.
Now return to the general theory of eigenvalue equations. Although I have never met the equation you wrote, I cannot find any place it can be wrong apart from the one just pointed out. Though, I don't see how far can you go from it.
Answer 3:
Hermite polynomials are usually beyond standard quantum mechanics courses. If you know Legendre, Chebyshev and/or other polynomials, you may guess that Hermite polynomials are derived as solution to some differential equation, and this does not contradict to the definition of $\psi$.
As I've already mentioned, Hermite polynomials are usually beyond standard quantum mechanics courses. Usually you are not supposed to derive them at this level. However, if you are still interested, you may want to consult with google or ask another question here.
Hope your questions have now been answered in full. However, should you need any further comment - you are welcome.
Let's start from
$$H = \hbar \omega \left(f^\dagger f - \frac{1}{2}\right),$$
with $\{f, f^\dagger\}=1$, $\{f, f\} = 0$ and define fermionic position and momentum coordinates by
$$ \psi_1 = \sqrt{\frac{\hbar}{2}} \left(f + f^\dagger\right) \\
\psi_2 = i\sqrt{\frac{\hbar}{2}} \left(f - f^\dagger\right) $$
with the following anticommutation relations:
$$
\{\psi_i, \psi_j\} = \hbar \delta_{ij}.$$ So the operators anticommute with each other and square to $\hbar/2$.
We the find the Hamiltonian formulated in the new coordinates
$$H = -i \omega \psi_1 \psi_2,$$
which clearly gives rise to oscillatory motion, as can be seen by calculating the Heisenberg equations of motion:
$$\dot \psi_1 = -\omega \psi_2 \\
\dot \psi_2 = +\omega\psi_1.
$$
This doesn't have the form you expected it to have, but that just shows the weirdness of fermionic degrees of freedom.
Best Answer
One straightforward and informative approach is to compare your 2 expressions for $a_H(t)$ in the basis of energy eigenstates. Consider the matrix element \begin{align} \langle m | e^{i H t} a e^{-iHt} | n\rangle &= e^{-i \omega (n-m)t}\langle m| a |n\rangle\;. \end{align} It is easy to see that this is zero unless $n = m+1$, in which case the exponential reduces to $e^{-i\omega t}$, so clearly this will equal $\langle m | e^{-\imath \omega t}a|n\rangle$ for all $m$ and $n$. Since the energy eigenstates form a complete basis, this implies they are equal as operators.
Having seen how the proof goes in a particular basis, it is easier to see how it must go in a basis free version. The key point is that the lowering operator acting on the state between the two exponentials creates an offset that results in the remaining exponential at the end. So we want to derive something along the lines of $$ e^{i\omega t n} a = a e^{i\omega t (n-1)} $$
With that in mind, first we write $$ na = an + [n,a] = a(n-1) $$ This implies that, for an analytic function $f$, \begin{align} f(n)a &= \sum_m \frac{ f^{(m)}(0)}{m!}n^m a\\ &= a \sum_m \frac{ f^{(m)}(0)}{m!}(n-1)^m\\ &= a f(n-1) \end{align} so that $$ e^{i \omega nt}ae^{-i\omega nt} = a e^{i \omega (n-1)t}e^{-i\omega nt} = a e^{-i\omega t} $$