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.
Best Answer
1) because the potential energy function has even symmetry, the solutions to Schrodinger's Equation results in purely even or odd solutions. Consider the Time Independent Schrodinger Equation (TISE) below
$$-\frac{\hbar^2}{2m}\nabla^2 \psi(x) + U(x)\psi(x) = E\psi(x)$$
But the potential contains the symmetry that $U(-x) = U(x)$. What this implies is that under the parity transform $$(x \implies -x)$$
$$-\frac{\hbar^2}{2m}\nabla^2 \psi(-x) + U(-x)\psi(-x) = E\psi(-x)$$ Or because of the potential's even symmetry
$$-\frac{\hbar^2}{2m}\nabla^2 \psi(-x) + U(x)\psi(-x) = E\psi(-x)$$
What this means is that both wavefunctions $\Psi(x)$ AND $\Psi(-x)$ solve this particular wave equation. Further, because the Schrodinger Equation is a linear Differential Equation, any linear combination of these two solutions must also be a solution to the equation. therefore, solutions exist such that:
$$\psi_{even}(x) = \psi(x) + \psi(-x)$$ $$\psi_{odd}(x) = \psi(x) - \psi(-x)$$
These follow from the definitions of even and odd functions which is easily verified.
2) Concerning the modulus squared, $\psi^* \psi$, if $\psi$ is always either even or odd, then when multiplying an odd function with itself you get an even function. likewise, an even function multiplied with itself will be even. Consider trivial polynomials for example,
$x^3$ is odd, but $(x^3)(x^3) = (x^3)^2 = x^6$ which is obviously even.
3) Lastly the fact that $\left<x \right> = 0$ follows from the previous statement, that $|\psi|^2$ is always even for potentials with even symmetry, like the SHO. This is because
$$\left<x \right> = \int_{- \infty} ^{\infty} x|\psi|^2dx$$
In the integrand, there is an odd function ($x$) times the definitely even modulus squared. This creates an odd integrand. For any odd integrand with symmetric limits, this integral must be zero.