Three. (50 points) $f(x)$ is a function defined on the set of natural numbers, satisfying
$$
f(1)=\frac{4}{3},
$$

and $2 f(k)=f(k-1)-k \cdot 5^{k}+2^{k}(k \geqslant 2, k \in \mathbf{N})$. Try to find the analytical expression of the function $f(x)$.
Given the known condition $2 f(k)=f(k-1)-k \cdot 5^{k}+2^{k}$, construct the following equation:
$$
2\left[f(k)-\frac{2^{k+1}}{3}\right]=\left[f(k-1)-\frac{2^{k}}{3}\right]-k \cdot 5^{k} .
$$

Divide both sides by $5^{k}$:
$$
\frac{2\left[f(k)-\frac{2^{k+1}}{3}\right]}{5^{k}}=\frac{1}{5} \cdot \frac{\left[f(k-1)-\frac{2^{k}}{3}\right]}{5^{k-1}}-k .
$$

Let $g(k)=\frac{f(k)-\frac{2^{k+1}}{3}}{5^{k}}$,
then $2 g(k)=\frac{1}{5} g(k-1)-k$
$$
\begin{array}{l}
\Rightarrow 2\left[g(k)+\frac{5 k}{9}\right]=\frac{1}{5}\left[g(k-1)+\frac{5(k-1)}{9}\right]+\frac{1}{9} \\
\Rightarrow 2\left[g(k)+\frac{5 k}{9}-\frac{5}{81}\right] \\
\quad=\frac{1}{5}\left[g(k-1)+\frac{5(k-1)}{9}-\frac{5}{81}\right] .
\end{array}
$$

Let $\varphi(k)=g(k)+\frac{5 k}{9}-\frac{5}{81}$,
then $2 \varphi(k)=\frac{1}{5} \varphi(k-1)$, i.e., $\varphi(k)=\frac{1}{10} \varphi(k-1)$.
Since $k$ is a natural number, and $f(1)=\frac{4}{3}$, from equations (1) and (2) we get
$$
g(1)=0, \varphi(1)=\frac{5}{9}-\frac{5}{81}=\frac{40}{81} \text {. }
$$

Thus, the first term of $\{\varphi(k)\}$ is $a_{1}=\varphi(1)=\frac{40}{81}$.
Also, $a_{k}=\varphi(k)=\frac{1}{10} \varphi(k-1)=a_{k-1} \cdot \frac{1}{10}$, i.e.,
$$
\frac{a_{k}}{a_{k-1}}=\frac{1}{10}=q \text {. }
$$

Therefore, $\{\varphi(k)\}$ is a geometric sequence with the first term $\frac{40}{81}$ and common ratio $\frac{1}{10}$. So, $\varphi(k)=\frac{40}{81}\left(\frac{1}{10}\right)^{k-1}$.

From equation (2), we get $g(k)=\varphi(k)-\frac{5 k}{9}+\frac{5}{81}$, i.e.,
$$
g(k)=\frac{40}{81} \cdot 10^{1-k}-\frac{5 k}{9}+\frac{5}{81} \text {. }
$$

From equation (11), we get $f(k)=5^{k} g(k)+\frac{2^{k+1}}{3}$, i.e.,
$$
f(k)=\frac{5^{k+1}}{81}\left(8 \cdot 10^{1-k}-9 k+1\right)+\frac{2^{k+1}}{3} \text {. }
$$

Thus, $f(n)=\frac{5^{n+1}}{81}\left(8 \cdot 10^{1-n}-9 n+1\right)+\frac{2^{n+1}}{3}$,
i.e., $f(x)=\frac{5^{x+1}}{81}\left(8 \cdot 10^{1-x}-9 x+1\right)+\frac{2^{x+1}}{3}$.