The struggle of the land, by 1996, the forest coverage rate of the whole county had reached 30% (becoming an oasis). From 1997 onwards, each year will see such changes: 16% of the original desert area will be converted, and 4% of it will be eroded, turning back into desert.
(1) Assuming the total area of the county is 1, in 1996 the oasis area is $a_{1}=\frac{3}{10}$, after $n$ years the oasis area becomes $a_{n+1}$. Prove:
$$
a_{n+1}=\frac{4}{5} a_{n}+\frac{4}{25} \text {. }
$$
(2) How many years at least are needed to make the forest coverage rate of the whole county exceed 60% (round to the nearest whole number)?
Let the accumulated area be $b_{n+1}$. Then
$$
a_{1}+b_{1}=1, \quad a_{n}+b_{n}=1 .
$$

According to the problem, $a_{n+1}$ consists of two parts: one part is the remaining area of the original oasis $a_{n}$ after being eroded by $\frac{4}{100} \cdot a_{n}$, which is
$$
a_{n}-\frac{4}{100} a_{n}=\frac{96}{100} a_{n} .
$$

The other part is the newly greened area $\frac{16}{100} b_{n}$. Therefore,
$$
\begin{aligned}
a_{n+1} & =\frac{96}{100} a_{n}+\frac{16}{100} b_{n}=\frac{96}{100} a_{n}+\frac{16}{100}\left(1-a_{n}\right) \\
& =\frac{4}{5} a_{n}+\frac{4}{25} .
\end{aligned}
$$
(2) From (1), we have
$$
\begin{aligned}
a_{n+1}-\frac{4}{5} & =\frac{4}{5}\left(a_{n}-\frac{4}{5}\right)=\left(\frac{4}{5}\right)^{2}\left(a_{n-1}-\frac{4}{5}\right) \\
& =\cdots=\left(\frac{4}{5}\right)^{n}\left(a_{1}-\frac{4}{5}\right) \\
& =-\frac{1}{2}\left(\frac{4}{5}\right)^{n} . \\
\therefore a_{n+1} & =\frac{4}{5}-\frac{1}{2}\left(\frac{4}{5}\right)^{n} .
\end{aligned}
$$

According to the problem, $a_{n+1} \geqslant 60 \%$, i.e.,
$$
\frac{4}{5}-\frac{1}{2}\left(\frac{4}{5}\right)^{n}>\frac{3}{5} \text {. }
$$

This simplifies to $\left(\frac{4}{5}\right)^{n}<\frac{2}{5}$,
or $(1-0.2)^{n}<0.4$.
To find the integer value, we use an upper approximation for $(1-0.2)^{n}$
$$
(1-0.2)^{n}<1-C_{n}^{1} \times 0.2+C_{n}^{2} \times 0.04 \leqslant 0.4,
$$

which simplifies to $n^{2}-11 n+30 \leqslant 0$.
Solving this, we get $5 \leqslant n \leqslant 6$.
Taking the smallest integer $n=5$, we verify that $a_{5}<0.6<a_{6}$, so after 5 years of effort, the greened area can exceed $60 \%$.

Here we used an approximation. We can also solve $a_{n} \leqslant 0.6<a_{n+1}$ to get $n=5$.