Five. (Full marks 20 points) A county is located in a desert area, and people have been engaged in a tenacious struggle against nature for a long time. By the end of 1998, the county's greening rate had reached $30 \%$. Starting from 1999, the following situation will occur every year: $16 \%$ of the original desert area will be greened, while at the same time, due to various reasons, $4 \%$ of the original greened area will be re-sandified.
(1) Let the total area of the county be 1, and the total greened area at the end of 1998 be $a_{1}=\frac{3}{10}$. After $n$ years, the total greened area is $a_{n+1}$. Prove: $a_{n+1} = \frac{4}{5} a_{n} + \frac{4}{25}$.
(2) How many years of effort are needed at least to achieve a county-wide greening rate?
(1) Let the current desert area be $b_{1}$, and after $n$ years, the desert area be $b_{n+1}$. Thus, $a_{1}+b_{1}=1, a_{n}+b_{n}=1$.

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 $\frac{4}{100} a_{n}$, which is $\frac{96}{100} a_{n}$; the other part is the newly greened area $\frac{16}{100} b_{n}$. Therefore,
$$
\begin{array}{l}
a_{n+1}=\frac{96}{100} a_{n}+\frac{16}{100} b_{n} \\
=\frac{96}{100} a_{n}+\frac{16}{100}\left(1-a_{n}\right)=\frac{4}{5} a_{n}+\frac{4}{25} .
\end{array}
$$
(2) From (1), we get
$$
\begin{array}{l}
a_{n+1}-\frac{4}{5}=\frac{4}{5}\left(a_{n}-\frac{4}{5}\right)=\left(\frac{4}{5}\right)^{2}\left(a_{n-1}-\frac{4}{5}\right) \\
=\cdots=\left(\frac{4}{5}\right)^{n}\left(a_{1}-\frac{4}{5}\right)=-\frac{1}{2} \times\left(\frac{4}{5}\right)^{n} .
\end{array}
$$

Therefore, $a_{n+1}=\frac{4}{5}-\frac{1}{2} \times\left(\frac{4}{5}\right)^{n}$.
Note that $a_{n+1} \geqslant 60 \%$, i.e., $\frac{4}{5}-\frac{1}{2} \times\left(\frac{4}{5}\right)^{n}>\frac{3}{5}$. Simplifying, we get $\left(\frac{4}{5}\right)^{n}\frac{1-2 \lg 2}{1-3 \lg 2}>4$. Therefore, it will take at least 5 years to make the county's greening rate exceed $60 \%$.