Three, (50 points) Find the smallest positive integer $k$, such that $625^{k} \equiv 1(\bmod 343)$
$$
\left(5^{4}\right)^{k} \equiv 1\left(\bmod 7^{3}\right),
$$

which can be transformed into $(89 \times 7+2)^{k}-1 \equiv 0\left(\bmod 7^{3}\right)$.
Expanding it, we get
$$
\begin{array}{l}
\mathrm{C}_{k}^{2}(89 \times 7)^{2} \times 2^{k-2}+\mathrm{C}_{k}^{1}(89 \times 7) \times 2^{k-1}+2^{k}-1 \\
\equiv 0\left(\bmod 7^{3}\right) .
\end{array}
$$

Thus, $7 \mid\left(2^{k}-1\right)$, i.e., $2^{k} \equiv 1(\bmod 7)$.
Given $2^{1} \equiv 2(\bmod 7), 2^{2} \equiv 4(\bmod 7), 2^{3} \equiv$ $1(\bmod 7)$, the order of 2 modulo 7 is 3, hence $3 \mid k$.
Let $k=3 s$. Substituting into equation (1), we get
$$
\begin{array}{l}
\mathrm{C}_{38}^{2}(89 \times 7)^{2} \times 2^{34-2}+\mathrm{C}_{35}^{1}(89 \times 7) \times 2^{34-1}+(7+1)^{5}-1 \\
\equiv 0\left(\bmod 7^{3}\right) .
\end{array}
$$

Then $\mathrm{C}_{3}^{2}(89 \times 7)^{2} \times 2^{3-2}+\mathrm{C}_{3}^{1}(89 \times 7) \times 2^{3-1}+$ $\mathrm{C}_{s}^{2} 7^{2}+\mathrm{C}_{s}^{1} 7 \equiv 0\left(\bmod 7^{3}\right)$.
Thus, $\mathrm{C}_{38}^{2} 89^{2} \times 7 \times 2^{3 s-2}+\mathrm{C}_{38}^{1} 89 \times 2^{34-1}+$ $\mathrm{C}_{3}^{2} 7^{2}+\mathrm{C}_{3}^{1}=0\left(\bmod 7^{2}\right)$.
Therefore, $3 s \times 89 \times 2^{3 s-1}+s \equiv 0(\bmod 7)$.
Also, $2^{3-1} \equiv 2^{3-3} \times 2^{2} \equiv\left(2^{3}\right)^{s-1} \times 2^{2} \equiv 2^{2}(\bmod 7)$, so $3 s \times 40 \times 2^{2}+s \equiv 0(\bmod 7)$, i.e., $481 s \equiv 0(\bmod 7)$.
Since $(481,7)=1$, it follows that $7 \mid s$.
Let $s=7 t$. Substituting into equation (2), we get
$$
\begin{array}{l}
\frac{21 t(21 t-1)}{2} \times 89^{2} \times 7 \times 2^{21 t-2}+21 t \times 89 \times 2^{21 t-1}+ \\
\frac{7 t(7 t-1)}{2} \times 7+7 t=0\left(\bmod 7^{2}\right) . \\
\text { Then } \frac{21 t(21 t-1)}{2} \times 89^{2} \times 2^{21 t-2}+3 t \times 89 \times 2^{21 t-1}+ \\
\frac{7 t(7 t-1)}{2}+t=0(\bmod 7) .
\end{array}
$$

Therefore, $3 t \times 89 \times 2^{21 t-1}+t \equiv 0(\bmod 7)$.
Also, $2^{21 t-1} \equiv 2^{21 t-3} \times 2^{2} \equiv\left(2^{3}\right)^{7 t-1} \times 2^{2}$ $=2^{2}(\bmod 7)$,
so $3 t \times 40 \times 2^{2}+t \equiv 0(\bmod 7)$, i.e., $481 t \equiv 0(\bmod 7)$.

Since $(481,7)=1$, it follows that $7 \mid t$.
Let $t=7 r$. Then $k=3 s=21 t=147 r \geqslant 147$. Therefore, the minimum value of $k$ is 147.