In this excercise they expect you to know about Bloch’s theorem already according to the answers, because how else are you supposed to know that there arises another factor (-1)^n in the periodic eigenstate when k=pi/a?
You don’t really need the Bloch’s theorem here. The factor (-1)^n comes directly from the plane wave ansatz in question 2:
\phi_n=\phi_0 e^{ikna}, \qquad \psi_n=\psi_0 e^{ikna}.
In question 7 we evaluate the wavefunction at
k=\frac{\pi}{a}.
Substituting this gives
e^{ikna}=e^{i\pi n}=(-1)^n,
because n is an integer. So neighboring unit cells have opposite signs: plus, minus, plus, minus, etc.
This is why the solution writes
|\Psi_\pm(k=\pi/a)\rangle
=
\frac{1}{\sqrt{2}}\sum_n (-1)^n
\left(|n,1\rangle \pm |n,2\rangle\right).
Bloch’s theorem is indeed a more universal way to get to this idea, but the plane-wave Ansatz used in this exercise is already enough.