quantum field theory - Doubts about spontaneous symmetry breaking

I have been exposed to the usual treatment about spontaneous symmetry breaking in the standard model but it shames me to admit that there are some loose ends I still have to tie up. For simplicity, instead of the standard model let's consider a $U(1)$ gauge theory with a complex scalar $\phi$ given by the Lagrangian

$$\mathcal{L}=|D_{\mu}\phi|^2-\frac{1}{4}(F_{\mu\nu})^2-V(\phi^*\phi)$$

The $V$ part is called the scalar potential and we take it to be

$$V=-\mu^2\phi^*\phi+\frac{\lambda}{2}(\phi^*\phi)^2$$

where both $\mu$ and $\lambda$ are positive and whose shape is the logo of this very site. It is straightforward to check that the minimums of the potential occur at the field value

$$\phi_0=\left(\frac{\mu^2}{\lambda}\right)^{1/2}$$

or at any other related to this one by the $U(1)$ symmetry $\phi_0=$

$$\phi_0=\left(\frac{\mu^2}{\lambda}\right)^{1/2}e^{i\alpha(x)}$$

Until here I have no problem. In the next step it is assumed that $\phi_0=\left(\frac{\mu^2}{\lambda}\right)^{1/2}$ is the vacuum expectation value (I will use the letter $v$ henceforth) of the field $\phi$. FIRST QUESTION. How does this follow? why does the minimum of the scalar potential give the vacuum expectation value of the field?

Be that as it may, we have that $\phi$ has a vacuum expectation value. The next step is to expand $\phi$ around its VEV

$$\phi=v+\psi$$

and by introducing this in the Lagrangian we get a massive gauge boson that eats a degree of freedom from $\phi$. My SECOND QUESTION is, why do we have to expand around the VEV of $\phi$ to get the spectrum of the theory?