1. The last four digits of the number $7^{355}$ are $\qquad$
II. 1.1943.
This problem is to find the remainder of $7^{355}$ divided by $10^{4}$.
$$
\begin{array}{l}
7^{355}=7 \times 49^{1 \pi}=7 \times 49 \times(50-1)^{2 \times 88} \\
=7 \times 49 \times(2400+1)^{28} \\
\equiv 7 \times 49 \times\left(1+2400 \mathrm{C}_{88}^{1}\right) \\
\equiv 343 \times[1+2400 \times(100-12)] \\
\equiv 343 \times(1-2400 \times 12) \equiv 343-343 \times 12 \times 2400 \\
\equiv 343-43 \times 12 \times 2400 \equiv 343-103200 \times 12 \\
\equiv 343-3200 \times 12 \equiv 343-38400 \\
\equiv 343-8400 \equiv-8057 \equiv 1943\left(\bmod 10^{4}\right) .
\end{array}
$$

Therefore, the last four digits of $7^{355}$ are 1943.