Three. (50 points)
The sequence of real numbers $a_{1}, a_{2}, \cdots a_{n}, \cdots$, is defined by the following equation:
$$
a_{n+1}=2^{n}-3 a_{n}, n=0,1,2, \cdots \text {. }
$$
1. Find the expression for $a_{n}$ in terms of $a_{0}$ and $n$;
2. Find $a_{0}$ such that for any positive integer $n, a_{n+1}>a_{n}$.
(3) (50 points)
(1) From $a_{n+1}=2^{n}-3 a_{n}$, we get $\frac{a_{n+1}}{(-3)^{n+1}}-\frac{a_{n}}{(-3)^{n}}=\frac{2^{n}}{(-3)^{n+1}}$,
Let $b_{n}=\frac{a_{n}}{(-3)^{n}}$, so $b_{n+1}-b_{n}=\frac{2^{n}}{(-3)^{n+1}}$,
Therefore, $b_{n}=b_{1}+\left(b_{2}-b_{1}\right)+\left(b_{3}-b_{2}\right)+\cdots+\left(b_{n}-b_{n-1}\right)$
$$
\begin{array}{l}
=b_{1}+\frac{2}{(-3)^{2}}+\frac{2^{2}}{(-3)^{3}}+\frac{2^{3}}{(-3)^{4}}+\cdots+\frac{2^{n-1}}{(-3)^{n}} \\
=b_{1}+\left(-\frac{1}{3}\right)\left[\left(-\frac{2}{3}\right)+\left(-\frac{2}{3}\right)^{2}+\cdots+\left(-\frac{2}{3}\right)^{n-1}\right] \\
=b_{I}+\left(-\frac{1}{3}\right) \frac{\left(-\frac{2}{3}\right)\left(1-\left(-\frac{2}{3}\right)^{n-1}\right)}{1-\left(-\frac{2}{3}\right)} \\
=b_{1}+\frac{2}{15}\left(1\left(\frac{2}{3}\right)^{n-1}\right),
\end{array}
$$

Therefore, $\frac{a_{x}}{(-3)^{n}}=\frac{a_{1}}{-3}+\frac{2}{15}\left(1-\left(-\frac{2}{3}\right)^{n-1}\right)$.
Therefore, $a_{n}=a_{1} \cdot(-3)^{n-1}+\frac{2}{15}\left[(-3)^{n}+3 \cdot 2^{n-1}\right]$
$$
\begin{array}{l}
=\left(1-3 a_{0}\right)(-3)^{n-1}+\frac{2}{15}\left[(-3)^{n}+3 \cdot 2^{n-1}\right] \\
=\frac{1}{5}\left[2^{n}+(-1)^{n-1} \cdot 3^{n}\right]+(-1)^{n} \cdot 3^{n} \cdot a_{0}
\end{array}
$$
(2)
$$
\begin{aligned}
a_{n+1}-a_{n}= & \frac{1}{5}\left[2^{n-1}+(-1)^{n} \cdot 3^{n+1}\right]+(-1)^{n+1} \cdot 3^{n+1} \cdot a_{0} \\
& -\frac{1}{5}\left[2^{n}+(-1)^{n-1} \cdot 3^{n}\right]-(-1)^{n} \cdot 3^{n} \cdot a_{0} \\
= & \frac{1}{5} \cdot 2^{n}+(-1)^{n} \cdot 4 \cdot 3^{n} \cdot\left(\frac{1}{5} \quad a_{0}\right)
\end{aligned}
$$

Therefore, $\frac{1}{3^{n}}\left(a_{n+1}-a_{n}\right)=\frac{1}{5}\left(\frac{2}{3}\right)^{n}+(-1)^{n} \cdot 4\left(\frac{1}{5}-a_{0}\right)$.
If $\frac{1}{5}-a_{0}>0$, using $\lim _{n \rightarrow \infty}\left(\frac{2}{3}\right)^{n}=0$, for very large odd $n$, we have $a_{n+1}-a_{n}a_{n}$, thus the required $a_{0}=\frac{1}{5}$.