= Solution
A <weakly stationary process> has finite second moments, a <mean> $m=E X_t$ independent of $t$, and a <covariance> depending only on the separation of the observations. Its <autocovariance function> and <autocorrelation function> are
$$
\gamma(h)=E[(X_{t+h}-m)(X_t-m)],\qquad \rho(h)=\gamma(h)/\gamma(0),
$$
where the <autocorrelation> requires $\gamma(0)>0$.
Write the <autoregressive polynomial> as $A(z)=1-\alpha z+\alpha z^2$, so the <autoregressive model> is $A(B)X_t=\epsilon_t$, with $B$ the <backshift operator>. It is important to distinguish a <weakly stationary process> from a <causal time series>: <weak stationarity> alone does not require $X_t$ to depend only on present and past <white noise>.
For a two-sided <weakly stationary process>, the <two-sided stationary inverse of an autoregressive polynomial> exists and is unique if $A$ has no root on the <unit circle>. To find the excluded parameters, set $z=e^{i\lambda}$. The imaginary part of $A(z)=0$ gives
$$
\alpha\sin\lambda(2\cos\lambda-1)=0.
$$
The case $\alpha=0$ is harmless. The possible frequencies are $0,\pi,\pm\pi/3$. Since $A(1)=1$, $A(-1)=1+2\alpha$ and $A(e^{\pm i\pi/3})=1-\alpha$, the exceptional parameters are $-1/2$ and $1$. Thus, with nondegenerate <white noise>, the literal two-sided answer is
$$
\boxed{\alpha\in\mathbb R\setminus\{-\tfrac12,1\}.}
$$
For completeness, away from those values $1/A(z)$ has an absolutely summable <Laurent series> on an annulus containing the <unit circle>. Its bilateral coefficients define an $L^2$-convergent <linear filter of a stationary time series> applied to $\epsilon_t$, giving the required solution. Applying the same inverse to the equation gives uniqueness. At an exceptional value, the <spectral measure of a stationary time series> would have to satisfy $|A(e^{i\lambda})|^2\mu_X(d\lambda)=\sigma^2d\lambda/(2\pi)$. A zero of $A$ on the <unit circle> makes the resulting density nonintegrable near that zero, contradicting finite <variance>.
If the usual additional <causal time series> convention is intended, the <causality root criterion for an autoregressive model> requires the roots of $A$ to lie outside the closed <unit disk>. Equivalently, the roots of $u^2-\alpha u+\alpha$ lie strictly inside it. The real quadratic stability conditions reduce to $|\alpha|<1$ and $1+2\alpha>0$, giving the narrower answer
$$
\boxed{-\tfrac12<\alpha<1\quad\text{for the causal solution}.}
$$
The printed question does not explicitly impose this extra convention. Both answers have therefore been distinguished. If $\sigma^2=0$, the exceptional parameters permit nonunique stationary homogeneous solutions rather than the nonexistence conclusion above.
At $\alpha=-1/12$,
$$
A(z)=(1-z/4)(1+z/3),\qquad
\frac1{A(z)}=\frac{3/7}{1-z/4}+\frac{4/7}{1+z/3}.
$$
Consequently the <Wold representation> is
$$
\boxed{X_t=\sum_{j=0}^{\infty}\psi_j\epsilon_{t-j},\qquad
\psi_j=\frac37\left(\frac14\right)^j+\frac47\left(-\frac13\right)^j.}
$$
The coefficients are absolutely summable, and $\psi_0=1$. The past of $X$ lies in the closed linear span of the past <white noise>; conversely $\epsilon_t=A(B)X_t$ recovers that noise from present and past $X$. Thus $\epsilon_t$ is orthogonal to the past of $X$ and is its one-step linear prediction error, as required for the <Wold decomposition>. There is no deterministic component.
Using orthogonality of the <white noise>, put $H=|h|$ and calculate the <autocovariance> by geometric sums:
$$
\begin{aligned}
\gamma(h)/\sigma^2
&=\sum_{j\geq0}\psi_j\psi_{j+H}\\
&=\frac9{49}\frac{(1/4)^H}{1-1/16}
+\frac{16}{49}\frac{(-1/3)^H}{1-1/9}
+\frac{12}{49}\frac{(1/4)^H+(-1/3)^H}{1+1/12}\\
&=\frac{192}{455}(1/4)^H+\frac{54}{91}(-1/3)^H.
\end{aligned}
$$
Hence
$$
\boxed{\gamma(h)=\frac{6\sigma^2}{455}\left[32(1/4)^{|h|}+45(-1/3)^{|h|}\right].}
$$
In particular, $\gamma(0)=66\sigma^2/65$ and $\gamma(1)=-6\sigma^2/65$; these also satisfy the <Yule-Walker equations> for this <autoregressive model>.
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