Question: learning QM and just have a few questions regarding normalizing wavefunctions.

1. Thus far, every initial wavefunction that we've normalized has had an undefined constant explicitly put out front (i.e., something like $\Psi (x,0) = A(a - x^{2})$, as an example) I was recently asked to normalize a wavefunction that didn't have any undetermined constant out front of the function involving x. Now it's okay for me to multiply the undefined constant, say $A$, to the wavefunction? Since, of course, the purpose of normalizing the wavefunction is finding a form such that the probability is 1 over all space.
2. If I'm given a complete wavefunction (i.e., a solution to the time dependent Schrodinger equation - $\Psi(x,t) = \psi(x)\phi(t)$ I can just perform the normalization at $t=0$ right? In other words, normalize the initial wavefunction $\Psi(x,0) = \psi(x)\phi(0)$. Since once a wavefunction is normalized, it remains normalized for all time, correct? Or do I need to perform the normalization at arbitrary time $t$ if given the complete wavefunction?
3. If I've just normalized a wavefunction and am determining the constant $A$ but have $|A|^{2}$ = $K$ where $K$ is just some constant then I necessarily have $A$ = $\pm\sqrt{K}$ but the problem makes no mention of the nature of the constant (i.e., doesn't say anything like "where $A,a,b,$ etc... are all positive real constants) then how am I to determine the sign of $A$? Does it even matter? I imagine it wouldn't in the verification of the normalization but it would change the sign of the wavefunction, so it must have some importance?
4. If I get a wavefunction such that $|\Psi(x,0)|^{2} = 1$ would that tell me that this wavefunction is non-normalizable? Since the integral doesn't converge.

Answer: 
1. Yes. Normalizing the wavefunction means multiplying it by whatever constant you need to in order to make its norm equal to $1$.
2. Yes, because time evolution in quantum mechanics is *unitary* - in other words, you can write $\psi(t) = \hat U(t) \psi(0)$ for a unitary operator $\hat U(t)$ called the *propagator*, and $\Vert \hat U \psi\Vert = \Vert \psi\Vert $ for all unitary operators $\hat U$. Caveat: this is no longer true if you perform a measurement, in which case your wavefunction will undergo (non-unitary) projective evolution and you may need to renormalize afterward.
3. If $|A|^2 = K$, then $A=e^{i\theta} \sqrt{K}$ for any arbitrary phase angle $\theta\in[0,2\pi)$. You may make any choice you want, and it will not impact any physically meaningful predictions of the theory. Perhaps you could check to see whether multiplying the wave function by some $e^{i\theta}$ changes e.g. the probabilities of any measurement outcomes.
1. yes
2. yes
3. the constant is determined up to an arbitrary phase. Remember that this is also the case for the wave function.