It is hard to know what exact the professor has done, but from what I have understood made an effort to help the situation.
Q1. A Gaussian function is chosen to represent a freely expanding wave function for several reasons:
(i) The Gaussian function represents a normal probability distribution function. Since $|\psi(x)|^2$ represents a probability distribution function for a huge class of particles moving in a similar way, and having momentum within a certain range, the large numbers theorem points towards a Gaussian function, to represent the wave function of particles within a region of space determined by the width, $\sigma$, of the Gaussian function. The $\sigma$ also happens to be the standard deviation, i.e. the uncertainty in the position of the particle.
(ii) The expectation value of the position of the particle turns out to be the value of $b$ in your Gaussian wave packet.
(iii) The Gaussian wave packet also contains the wavy bit shown by the phase factor
$e^{ikx}$.
This is what makes Gaussian wave packets an excellent representation of particles, which are known to be located within some region of width $w$, but nevertheless they are all moving as plane waves while the packet travels along in space.
(iv) The wave packet is made of an infinitely large number of momentum values in a momentum width which relates to $\sigma$ by a Fourier transformation.
Q2. So to put some order in all these, let us consider the general Gaussian function
$\psi(x)=\psi_0e^{-A(x-x_0)^2+Bx+C}$
which has a Fourier transformation
$\psi(k)=\psi_0\sqrt{\frac{\pi}{A}}e^{\frac{B^2}{4A}+Bx_0+C}$.
The normalisation constant $\psi_0$ is given by an integration in the range [-$\infty, +\infty$] and has the value $\sqrt{\frac{A}{\pi}}$ but leave it as $\psi_0$.
Q3. To make contact with your professor, try to apply these results using the following:
$A=\frac{1}{2\sigma^2}$
$B=ik$
$x_0=b$
$C=0$
The Fourier-transform operators
$$
\hat F = \int \mathrm dx \frac{e^{ikx}}{\sqrt{2\pi}}
\qquad
\hat F{}^{-1} = \int \mathrm dk \frac{e^{-ikx}}{\sqrt{2\pi}}
$$
are prettier if you attach a $\sqrt{2\pi}$ to them both, instead of the asymmetric
$$
\hat F_\text{ugly} = \int \mathrm dx \ {e^{ikx}}
\qquad
\hat F{}_\text{ugly}^{-1} = \int \mathrm dk \frac{e^{-ikx}}{{2\pi}}
$$
You should think of $\Psi(x) = \hat F {}^{-1} A(k)$ as your wavefunction, but $A(k) = \hat F\Psi(x)$ as also your wavefunction, in momentum space. You’re currently stuck because the normalization of $\Psi$ is related to the normalization of $A$.
Perhaps abetting your confusion: in the current version of your question (v3), you erroneously have $\Psi(x) = \hat F A(k)$. That’s impossible, because the integral on the right side eats up the dummy variable $x$, so the left side should be a function of $k$.
Likewise you have to be careful about the dummy variables when you are computing your overlap integrals. You expect that
\begin{align}
1 = \left< \Psi | \Psi \right>
&= \int \mathrm dx\ \Psi^*(x) \Psi(x)
\\
&= \int \mathrm dx
\left(
\int\mathrm dk A^*(k) \frac{e^{+ikx}}{\sqrt{2\pi}}
\right)
\left(
\int\mathrm dk’ A(k’) \frac{e^{-ik’x}}{\sqrt{2\pi}}
\right)
\end{align}
You’re going to exploit
$ \int\mathrm dx\ e^{i(k-k’)x} \sim \delta(k-k’)
$
and get rid of two of the integrals, leaving you with
$$
1= \left<\Psi|\Psi\right>_x = \left<A|A\right>_k
$$
which is what I meant about the normalizations being related. But you’re going to look up and/or prove that there are the correct amount of $\sqrt{2\pi}$ involved, rather than trusting my one-squiggle relationship.
If you have been putting off reading about Fourier transforms, today is a good day.
Best Answer
The expression $|\psi(x)|^2$ is the complex modulus squared; $$ |\psi(x)|^2 = \psi(x)^*\psi(x) $$ where here the star means complex conjugation. It follows that for any wavefunction of the form $$ \psi(x) = Ae^{ikx} f(x) $$ where $A$ and $f(x)$ are real, one has $$ |\psi(x)|^2 = A^2 f(x)^2 $$ since $|e^{ikx}|^2 = e^{ikx} e^{-ikx} = 1$.