Take a plane wave:
$$ A(x, t) = Ae^{i\phi(x, t)} = Ae^{i(kx-\omega t)}$$
where
$$ \phi(x, t) = kx-\omega t$$
is the phase. The time derivative of the phase,
$$ \frac{\partial \phi}{\partial t} = -\omega $$
gives the frequency. Meanwhile, the spatial derivative
$$ \frac{\partial \phi}{\partial x} = k $$
is the wavenumber:
$$ k = \frac{2\pi}{\lambda} $$
The phase velocity is their ratio:
$$ v_{ph} = \frac{\omega} k $$
So that answer to (1) is yes.
In the above wave:
$$ \omega(k) = v_{ph} k$$
This is called the dispersion relation, or: how does the frequency depend on wavenumber?
The linear relation is called "dispersionless": all frequencies propagate at the same speed.
For a non-linear relation, such as:
$$ \omega(k) = \sqrt{(ck)^2 + (mc^2)^2} $$
then:
$$ v_{ph} = \frac{\sqrt{(ck)^2 + (mc^2)^2}} k = c\sqrt{1+\frac{m^2c^4}{k^2}} > c$$
which is larger than $c$. So the answer to (2) is "yes".
You should convince yourself that the phase $\phi(x, t)$ is local, that is, as it changes and say, the peak, or the zero crossing, moves: no information is being transferred (really: nothing more than an apparent position is "moving").
Energy (or information) travels at the group velocity:
$$ v_{gp} = \frac{d\omega}{dk} $$
which in the example given, is:
$$ v_{gp} = c\frac k {\omega} < c$$
