1. Container $A$ contains 6 liters of saltwater with a concentration of $a \%$, and container $B$ contains 4 liters of saltwater with a concentration of $b \%$. After pouring 1 liter of solution from $A$ into $B$ and mixing, then pouring 1 liter of solution from $B$ back into $A$, this process is repeated $k$ times (pouring from $A$ to $B$, then from $B$ to $A$ counts as one repetition). After $k$ repetitions, the concentrations of the saltwater in $A$ and $B$ are $a_{k} \%$ and $b_{k} \%$ respectively.
（1）Find the sequence formed by $b_{k}-a_{k}$;
（2）Find $a_{k}$ and $b_{k}$.
1. Solution: (1) Let the concentrations of $A$ and $B$ after $k-1$ operations be $a_{k-1} \%$ and $b_{k-1} \%$ respectively, where $k \geqslant 1$, $a_{0}=a$, and $b_{0}=b$. According to the problem, we have
$$
\begin{array}{l}
b_{k}=\frac{1}{5}\left(4 b_{k-1}+a_{k-1}\right)=\frac{1}{5} a_{k-1}+\frac{4}{5} b_{k-1} \\
a_{k}=\frac{1}{6}\left(5 a_{k-1}+b_{k}\right)=\frac{13}{15} a_{k-1}+\frac{2}{15} b_{k-1}
\end{array}
$$
Subtracting (2) from (1) gives
$$
b_{k}-a_{k}=-\frac{10}{15} a_{k-1}+\frac{10}{15} b_{k-1}=\frac{2}{3}\left(b_{k-1}-a_{k-1}\right)
$$

Thus, the sequence $\left\{b_{k}-a_{k}\right\}$ is a geometric sequence with the first term $b_{0}-a_{0}=b-a$ and common ratio $\frac{2}{3}$.
(2) From the above result, we have
$$
b_{k}-a_{k}=(b-a)\left(\frac{2}{3}\right)^{k}
$$

From (1), we have
$$
b_{k+1}=\frac{1}{5} a_{k}+\frac{4}{5} b_{k}
$$

Eliminating $a_{k}$ using (3) and (4) gives
$$
b_{k+1}-b_{k}=\frac{1}{5}(a-b)\left(\frac{2}{3}\right)^{k}
$$

Therefore, $b_{k}=b_{0}+\sum_{i=0}^{k-1}\left[\frac{1}{5}(a-b)\left(\frac{2}{3}\right)^{i}\right]=b+\frac{1}{5}(a-b) \frac{1-\left(\frac{2}{3}\right)^{k}}{1-\frac{2}{3}}$
Thus, $b_{k}=\frac{3 a+2 b}{5}-(a-b) \cdot \frac{3}{5}\left(\frac{2}{3}\right)^{k}$
This formula still holds when $k=0$.
From (3), we get
$$
a_{k}=\frac{3 a+2 b}{5}+(a-b) \frac{2}{5}\left(\frac{2}{3}\right)^{k} .
$$