Since $k_E \propto \hat{p}^2$ and $\hat{p}$ is Hermitian you may see that this makes $k_E$ positive semidefinite, that is all of its eigenvalues are larger or equal to 0.
In other words when you measure this operator you will always get results which are larger or greater than zero.
This "contradiction" is resolved by the fact that the potential is a function of the coordinate and the momentum (and hence kinetic energy) does not commute with functions of coordinate. This in turn means that you can't simultaneously measure both $k_E$, $E$, and potential.
1) A time dependent state can be written as a superposition of the eigenstates of the Hamiltonian. I am assuming that the Hamiltonian is a constant one; does not change with time. The coefficients or weights of the eigenstates are complex.
so, for any state $\psi$,
$\psi(t) = c_1 \phi_1 + c_2 \phi_2 + \dots $
where $\phi_i$ are the normalized eigenstates of the Hamiltonian. The coefficients
$c_i(t) = \langle \phi_i|\psi(t) \rangle$
are time dependent.
And the time dependence has the precise form:
$c_i(t)=\exp(-\imath E_i t/\hbar) c_i(0)$
where $E_i$ is the energy of the state $\phi_i$. Thus given the initial coefficients $c_i(0)$, you can determine the coefficients $c_i(t)$ and therefore the state at any other instant $t$. Equivalently you can take $\psi(0)$ also as the initial condition, as $c_i(0)$ can be inferred from $c_i(0)=\langle \phi_i|\psi(0) \rangle$.
2) Constant energy does not mean a constant momentum. This is because the box is not translation invariant. However, the magnitude of the momentum is a constant (in the case of the simple 1D quantum well. This is not true in general). You can in fact write the energy eigenstate as a linear combination of states of momenta $p$ and $-p$. So momentum distribution is nonzero at $\pm p$ only.
3) Distribution of an observable can be interpreted as follows. The probability of finding a value of $o$ for a measurement (say position or momentum) on a system in $\psi$. Such measurements are associated with a corresponding operator $O$.
$p(o) = |\langle\psi | \phi_o\rangle|^2$ if $o$ is an eigenvalue of $O$, and $p(o)=0$ otherwise.
$\phi_o$ is the normalized eigenstate of $O$ withe eigenvalue $o$. (I am assume no degeneracy).
Best Answer
The walls do not really matter that much here. I would say that the only points that matter to understand that quote are:
These two conditions are enough to say that the total energy of the particle (its value) is the kinetic energy it has in a big fraction of the box.