8・33 The sequence of real numbers $a_{1}, a_{2}, \cdots, a_{n}, \cdots$ is defined by the following equation:
$$a_{n+1}=2^{n}-3 a_{n}, n=0,1,2, \cdots$$
(1) Find the expression for $a_{n}$ in terms of $a_{0}$ and $n$.
(2) Find $a_{0}$ such that for any positive integer $n, a_{n+1}>a_{n}$.
[Solution] (1) According to the definition of the sequence, we have
$$\begin{array}{l}
a_{n}=2^{n-1}-3 a_{n-1} \\
-3 a_{n-1}=-3 \cdot 2^{n-2}+3^{2} a_{n-2} \\
3^{2} a_{n-2}=3^{2} \cdot 2^{n-3}-3^{3} a_{n-3} \\
\cdots \cdots \cdots \cdots \\
(-1)^{n-1} 3^{n-1} a_{1}=(-1)^{n-1} 3^{n-1} 2^{0}+(-1)^{n} \cdot 3^{n} a_{0}
\end{array}$$

Adding all these equations, we get
$$\begin{aligned}
a_{n}= & 2^{n-1}-3 \cdot 2^{n-2}+3^{2} \cdot 2^{n-3}-\cdots+(-1)^{n-1} \cdot 3^{n-1} \cdot 2^{0}+ \\
& (-1)^{n} \cdot 3^{n} a_{0} \\
= & \frac{2^{n-1} \cdot\left[1-\left(-\frac{3}{2}\right)^{n}\right]}{1-\left(-\frac{3}{2}\right)}+(-1)^{n} \cdot 3^{n} a_{0} \\
= & \frac{1}{5}\left[2^{n}+(-1)^{n-1} 3^{n}\right]+(-1)^{n} \cdot 3^{n} \cdot a_{0}
\end{aligned}$$
(2)
$$\begin{aligned}
a_{n+1}-a_{n}= & \frac{1}{5}\left[2^{n+1}+(-1)^{n} \cdot 3^{n+1}\right]+(-1)^{n+1} \cdot 3^{n+1} \cdot a_{0} \\
& -\frac{1}{5}\left[2^{n}+(-1)^{n-1} \cdot 3^{n}\right]-(-1)^{n} \cdot 3^{n} a_{0} \\
= & \frac{1}{5}\left[2^{n}+(-1)^{n} \cdot 4 \cdot 3^{n}\right]+(-1)^{n+1} \cdot 4 \cdot 3^{n} \cdot \\
& a_{0} \\
= & \frac{1}{5} \cdot 2^{n}+(-1)^{n} \cdot 4 \cdot 3^{n}\left(\frac{1}{5}-a_{0}\right)
\end{aligned}$$

Then $\frac{1}{3^{n}}\left(a_{n+1}-a_{n}\right)=\frac{1}{5} \cdot\left(\frac{2}{3}\right)^{n}+(-1)^{n} \cdot 4 \cdot\left(\frac{1}{5}-a_{0}\right)$, from the above equation, if $\frac{1}{5}-a_{0} \neq 0$, there exists a sufficiently large $n$ (odd or even), which can make $a_{n+1}-a_{n}0$ for any positive integer $n$.

Therefore, the required $a_{0}=\frac{1}{5}$.