= Solution
Take $\alpha^2+\beta^2=1$ and label the first vector of each measurement basis by outcome $0$. Assign the <Pauli measurement> value $+1$ to outcome $0$ and $-1$ to outcome $1$. The four observables are
$$
A_1=Z,\qquad A_2=X,\qquad
B_1=\frac{\sqrt3}{2}X+\frac12 Z,\qquad
B_2=\frac{\sqrt3}{2}X-\frac12 Z.
$$
Here $X$ and $Z$ are the <Pauli X gate> and <Pauli Z gate> matrices. These follow by subtracting the two rank-one basis projectors; a real basis rotated through $\gamma$ has observable $\sin(2\gamma)X+\cos(2\gamma)Z$.
The <Schmidt-basis Pauli correlation tensor> of the state gives
$$
\langle Z\otimes Z\rangle=1,\quad \langle X\otimes X\rangle=2\alpha\beta,\quad
\langle X\otimes Z\rangle=\langle Z\otimes X\rangle=0.
$$
Consequently $E_{11}=1/2$, $E_{21}=E_{22}=\sqrt3\alpha\beta$ and $E_{12}=-1/2$, where $E_{jk}=\langle A_j\otimes B_k\rangle$. For binary outcomes, $\mathbb E\big([A-B]_2\big)=P(A\ne B)=(1-E_{AB})/2$, whereas the final offset term has $\mathbb E\big([B-A-1]_2\big)=P(A=B)=(1+E_{AB})/2$. The <chained modular Bell inequality> left side is therefore
$$
I=\frac{1-E_{11}}2+\frac{1-E_{21}}2+\frac{1-E_{22}}2+\frac{1+E_{12}}2
=\boxed{\frac32-\sqrt3\alpha\beta}.
$$
The local bound is $I\geq1$, so the exact violation condition is
$$
\boxed{\alpha\beta>\frac1{2\sqrt3},\qquad \alpha^2+\beta^2=1}.
$$
Equality saturates the bound. The maximal violation for these fixed measurements occurs at $\alpha=\beta=1/\sqrt2$ or their common negative, giving $I=(3-\sqrt3)/2$. Opposite signs do not violate this particular inequality with these fixed bases, although other measurement choices can reveal the state's <entanglement>. If unnormalized real amplitudes are used, replace $\alpha\beta$ throughout by $\alpha\beta/(\alpha^2+\beta^2)$.
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