= Solution
<Bloch theorem> states that the energy eigenstates of a Hamiltonian invariant under translations by a <Bravais lattice> can be chosen as <Bloch states>
$$
\boxed{
T_r|\psi_k\rangle=e^{-ik\cdot r}|\psi_k\rangle,
\qquad
\psi_k(x)=e^{ik\cdot x}u_k(x),
\qquad
u_k(x+r)=u_k(x).}
$$
Indeed, the <unitary operators> $T_r$ commute with one another because $T_rT_s=T_{r+s}=T_sT_r$, and they commute with $H$ by hypothesis. The <simultaneous diagonalization> theorem therefore lets us diagonalize all translations within each energy eigenspace. Their eigenvalues form a unitary character $\chi$ of the additive lattice:
$$
\chi(r+s)=\chi(r)\chi(s),
\qquad |\chi(r)|=1.
$$
Writing $\chi(a_j)=e^{-ik\cdot a_j}$ on a primitive basis gives $\chi(r)=e^{-ik\cdot r}$. In position space, with $(T_r\psi)(x)=\psi(x-r)$, this implies $\psi_k(x+r)=e^{ik\cdot r}\psi_k(x)$. Hence $u_k(x)=e^{-ik\cdot x}\psi_k(x)$ is lattice-periodic. Adding a <reciprocal lattice> vector to $k$ leaves the character unchanged, so the <crystal momentum> lies in a <Brillouin zone>.
For
$$
a_1=\frac a2(\sqrt3,1),
\qquad
a_2=\frac a2(\sqrt3,-1),
$$
the equations $a_i\cdot b_j=2\pi\delta_{ij}$ give the reciprocal basis
$$
\boxed{
b_1=\left(\frac{2\pi}{\sqrt3a},\frac{2\pi}{a}\right),
\qquad
b_2=\left(\frac{2\pi}{\sqrt3a},-\frac{2\pi}{a}\right).}
$$
The reciprocal lattice is triangular, so the first Brillouin zone is its <Wigner-Seitz cell>, a regular hexagon. Its six corners are
$$
\pm\frac{2b_1+b_2}{3},
\qquad
\pm\frac{b_1+2b_2}{3},
\qquad
\pm\frac{b_1-b_2}{3}.
$$
Reciprocal-lattice translations identify these corners in two classes of three. Representatives are
$$
\boxed{
K=\frac{2b_1+b_2}{3}
=\left(\frac{2\pi}{\sqrt3a},\frac{2\pi}{3a}\right),
\qquad
K'=\frac{b_1+2b_2}{3}
=\left(\frac{2\pi}{\sqrt3a},-\frac{2\pi}{3a}\right).}
$$
For example, $(b_1-b_2)/3=K'-b_2$, while the other equivalences follow by symmetry and reciprocal translations. This gives the requested sketch: a regular hexagon with the vertical edge from $K'$ to $K$ at $k_x=2\pi/(\sqrt3a)$ and alternating $K,K'$ corner classes.
For the <tight-binding model>, introduce the normalized Bloch sum
$$
|k\rangle=\frac1{\sqrt{N_s}}\sum_{r\in\Lambda}e^{ik\cdot r}|r\rangle.
$$
Each hop by $\pm a_j$ multiplies this state by $e^{\mp ik\cdot a_j}$. Thus the <two-direction nearest-neighbour tight-binding dispersion> is
$$
\boxed{
\begin{aligned}
E(k)
&=E_0-2\lambda\{\cos(k\cdot a_1)+\cos(k\cdot a_2)\}\\
&=E_0-4\lambda
\cos\left(\frac{\sqrt3ak_x}{2}\right)
\cos\left(\frac{ak_y}{2}\right).
\end{aligned}}
$$
Along the boundary edge from $K'$ to $K$, $k_x=2\pi/(\sqrt3a)$ and $|k_y|\leq2\pi/(3a)$, so
$$
E(k)=E_0+4\lambda\cos\left(\frac{ak_y}{2}\right).
$$
For $\lambda>0$, it rises from $E_0+2\lambda$ at either corner to $E_0+4\lambda$ at the midpoint; for $\lambda<0$ the ordering reverses. Globally the two cosines in the first expression can simultaneously equal $1$ or $-1$, and therefore
$$
\boxed{
E_{\min}=E_0-4|\lambda|,
\qquad
E_{\max}=E_0+4|\lambda|,
\qquad
\text{band width}=8|\lambda|.}
$$
Each orbital band contains two one-electron states per lattice site because an electron has two spin states. By <band filling>, a valency of one leaves this band half-filled and the material conducts, whereas a valency of two fills it. A filled band can be insulating only if it is separated from every empty band by a positive <band gap>.
The first band's maximum is $E_0+4|\lambda|$. If the second band's minimum is $E_0+\Delta$, a gap exists precisely when $\Delta>4|\lambda|$. Consequently, among the nontrivial fillings described here,
$$
\boxed{\text{the material is a band insulator when the valency is }2
\text{ and }\Delta>4|\lambda|.}
$$
For valency one the first band is partially filled. If $\Delta\leq4|\lambda|$, the bands overlap or touch, so even at valency two the <overlapping energy bands prevent a band insulator>.
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