= Solution
Set
$$
p_n=\mathbb P(X_n=1\mid X_0=1).
$$
Conditioning on $X_n$ gives the affine recurrence
$$
p_{n+1}=(1-\alpha)p_n+\beta(1-p_n)
=\beta+(1-\alpha-\beta)p_n.
$$
Its fixed point is the first component of the <stationary distribution>,
$$
\pi_1=\frac{\beta}{\alpha+\beta}.
$$
Thus $p_n-\pi_1=(1-\alpha-\beta)^n(p_0-\pi_1)$, and $p_0=1$ gives
$$
\boxed{\mathbb P(X_n=1\mid X_0=1)
=\frac{\beta}{\alpha+\beta}
+\frac{\alpha}{\alpha+\beta}(1-\alpha-\beta)^n}.
$$
Starting instead from state two gives the same recurrence with initial value zero. Therefore
$$
\boxed{\mathbb P(X_n=1\mid X_0=2)
=\frac{\beta}{\alpha+\beta}
\left[1-(1-\alpha-\beta)^n\right]}.
$$
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