8. As shown in the figure, in a triangular pond divided into 9 regions, a frog starts at region $A$. Every minute, the frog randomly jumps to an adjacent (Note: triangles with a common side are called adjacent) region. What is the probability that the frog returns to $A$ after 2022 minutes? $\qquad$
Re-label the regions as shown in the figure. Since the frog jumps to a different color region each time, after an odd number of minutes, the frog will be in the black shaded regions $a, b, c$. Let the probabilities of the frog being in regions $a, b, c$ after $2k-1$ minutes be $p_{k}, q_{k}, q_{k}$, respectively, so $p_{k}+2 q_{k}=1$. The two-jump scenarios from region $a$ are $a 1 a, a 1 a, a 2 a, a 2 b, a 3 a, a 3 c$; from region $b$ are $b 2 a, b 2 b, b 4 b, b 4 b, b 5 b, b 5 c$; from region $c$ are $c 3 a, c 3 c, c 5 b, c 5 c, c 6 c, c 6 c$. Each of these scenarios has a probability of $\frac{1}{6}$. Thus, $p_{k+1}=\frac{2}{3} p_{k}+\frac{1}{6} q_{k}+\frac{1}{6} q_{k}=\frac{2}{3} p_{k}+\frac{1}{3} q_{k}, q_{k+1}=\frac{1}{6} p_{k}+\frac{2}{3} q_{k}+\frac{1}{6} q_{k}=\frac{1}{6} p_{k}+\frac{5}{6} q_{k}$. Using $p_{k}+2 q_{k}=1 \Rightarrow p_{k+1}=\frac{2}{3} p_{k}+\frac{1}{3} q_{k}=\frac{2}{3} p_{k}+\frac{1}{6}\left(1-p_{k}\right)=\frac{1}{2} p_{k}+\frac{1}{6}$ $\Rightarrow p_{k+1}-\frac{1}{3}=\frac{1}{2}\left(p_{k}-\frac{1}{3}\right)$, combined with $p_{1}=1$, we get $p_{k}=\frac{2}{3} \cdot \frac{1}{2^{k-1}}+\frac{1}{3}\left(k \in \mathbf{N}^{*}\right)$. After 2021 minutes, the probability of the frog being in region $a$ is $p_{1011}=\frac{2}{3} \cdot \frac{1}{2^{1010}}+\frac{1}{3}$, so after 2022 minutes, the probability of the frog returning to $A$ is $P=\frac{2}{9} \cdot \frac{1}{2^{1010}}+\frac{1}{9}$.