Example 9 In the sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$, $a_{1}=b_{1}=1$, and $\left\{\begin{array}{l}a_{n+1}=b_{n}+3 \cdots \text { (1) } \\ b_{n+1}=2 a_{n}-1 \cdots \text { (2) }\end{array}\left(n \in \mathbf{N}^{*}\right)\right.$, find $a_{n}$ and $b_{n}$.
Solution: First, find the recurrence formulas for $\left\{a_{n}\right\},\left\{b_{n}\right\}$:
From (1), we get $a_{n+2}=b_{n+1}+3$.
Substitute (2) into the above equation to eliminate $b_{n+1}$, we get
$$
a_{n+2}=\left(2 a_{n}-1\right)+3=2 a_{n}+2 \Rightarrow a_{n+2}=2 a_{n}+2\left(n \in \mathbf{N}^{*}\right)
$$

Similarly, $b_{n+2}=2 b_{n}+5\left(n \in \mathbf{N}^{*}\right) \cdots$ (4)
Next, use (3) to find the general term of the sequence $\left\{a_{n}\right\}$:
In (3), let $x_{n}=a_{2 n-1}, y_{n}=a_{2 n}\left(n \in \mathbf{N}^{*}\right)$, then
$$
x_{1}=a_{1}=1, y_{1}=a_{2}=b_{1}+3=4 \text {, and }\left\{\begin{array}{l}
x_{n+1}=2 x_{n}+2 \\
y_{n+1}=2 y_{n}+2
\end{array}\left(n \in \mathbf{N}^{*}\right)\right. \text {. }
$$

These are two geometric difference sequences, which can be written as follows:
$$
x_{n+1}+2=2\left(x_{n}+2\right), y_{n+1}+2=2\left(y_{n}+2\right)\left(n \in \mathbf{N}^{*}\right) \text {. }
$$

From this, we get
$$
x_{n}+2=\left(x_{1}+2\right) \cdot 2^{n-1}=3 \cdot 2^{n-1}, y_{n}+2=\left(y_{1}+2\right) \cdot 2^{n-1}=6 \cdot 2^{n-1}=3 \cdot 2^{n} \text {. }
$$

Thus, $x_{n}=3 \cdot 2^{n-1}-2 ; y_{n}=3 \cdot 2^{n}-2\left(n \in \mathbf{N}^{*}\right)$.
Therefore, $a_{2 n-1}=3 \cdot 2^{n-1}-2, a_{2 n}=3 \cdot 2^{n}-2$.
So, $a_{n}=\left\{\begin{array}{ll}3 \cdot 2^{\frac{n-1}{2}}-2 & (2 \times n) \\ 3 \cdot n^{\frac{n}{2}}-2 & (2 \mid n)\end{array}\left(n \in \mathbf{N}^{*}\right)\right.$.
Or, it can be written as
$$
a_{n}=\frac{1-(-1)^{n}}{2}\left(3 \cdot 2^{\frac{n-1}{2}}\right)+\frac{1+(-1)^{n}}{2}\left(3 \cdot 2^{\frac{n}{2}}-2\right)\left(n \in \mathbf{N}^{*}\right) \text {. }
$$

Next, use (4) to find the general term of the sequence $\left\{b_{n}\right\}$:
In (4), let $x_{n}=b_{2 n-1}, y_{n}=b_{2 n}\left(n \in \mathbf{N}^{*}\right)$, then
$$
x_{1}=b_{1}=1, y_{1}=b_{2}=2 a_{1}-1=1 \text {, and } x_{n+1}=2 x_{n}+5, y_{n+1}=2 y_{n}+5\left(n \in \mathbf{N}^{*}\right) \text {. }
$$

These are two geometric difference sequences, which can be written as
$$
x_{n+1}+5=2\left(x_{n}+5\right), y_{n+1}+5=2\left(y_{n}+5\right)\left(n \in \mathbf{N}^{*}\right) \text {. }
$$

From this, we get
$$
x_{n}+5=\left(x_{1}+5\right) \cdot 2^{n-1}=3 \cdot 2^{n} ; y_{n}+5=\left(y_{1}+5\right) \cdot 2^{n-1}=3 \cdot 2^{n} \text {. }
$$

So, $x_{n}=3 \cdot 2^{n}-5 ; y_{n}=3 \cdot 2^{n}-5\left(n \in \mathbf{N}^{*}\right)$.
Therefore, $b_{2 n-1}=3 \cdot 2^{n}-5, b_{2 n}=3 \cdot 2^{n}-5\left(n \in \mathbf{N}^{*}\right)$.
So, $b_{n}=\left\{\begin{array}{ll}3 \cdot 2^{\frac{n+1}{2}}-5 & (2 \times n) \\ 3 \cdot 2^{\frac{n}{2}}-5 & (2 \mid n)\end{array} \quad\left(n \in \mathbf{N}^{*}\right)\right.$.
Or, it can be written as
$$
b_{n}=\frac{1-(-1)^{n}}{2}\left(3 \cdot 2^{\frac{n+1}{2}}-5\right)+\frac{1+(-1)^{n}}{2}\left(3 \cdot 2^{\frac{n}{2}}-5\right)\left(n \in \mathbf{N}^{*}\right) .
$$