if the potential $V(x)=V(-x)$ (is even), then $\psi(x)$ can be taken as even or odd
$\displaystyle -\frac{\hbar^{2}}{2m}\frac{d^{2}\psi(x)}{dx^{2}}+V(x)\psi(x)=E\psi(x)$ is the same as $\displaystyle -\frac{\hbar^{2}}{2m}\frac{d^{2}\psi(-x)}{dx^{2}}+V(x)\psi(-x)=E\psi(-x)$
linear combinations for even: $\psi(x) + \psi(-x)$ = even for odd $\psi(x) - \psi(-x)$ = odd
from now on I get confused, suppose we have a solution $\psi$ as an odd function, does that mean it could get minus value, but than $E\psi(-x)=-E\psi(-x)$ and this is nonsense.
mathematically no matter what funcion you insert you get the same result as stated before, but intuitively I am lost here ;]
Another part what if function is nor even nor odd?
The wavefunction $\psi$ generally takes complex values, so it's not immediately clear what a "minus" value would be – do you mean a negative real value? If so: Yes, any wavefunction, not just an even or odd one, can take negative real values. Since you can multiply any solution of the Schrödinger equation by an arbitrary phase factor, you can make it take negative real values wherever you like. It's not clear to me why you write that this implies $E\psi(-x)=-E\psi(-x)$.
As to the second part of your question: There may indeed be a solution that is neither even, nor odd (this would be called an "accidental degeneracy"). In this case you can take the linear combinations that you described, $\psi(x)+\psi(-x)$ and $\psi(x)-\psi(-x)$, to obtain an even solution and an odd solution.