II. Find the last four digits of $2^{1999}$.

Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly.
$$
\varphi(625)=\varphi\left(5^{4}\right)=5^{4} \times\left(1-\frac{1}{5}\right)=500 \text {. }
$$

Since $(2,625)=1$, by Euler's theorem, we get $2^{500} \equiv 1(\bmod 625)$.
$$
\begin{array}{l}
\therefore 2^{2000} \equiv 1(\bmod 625) . \\
\therefore \text { Let } 2^{2000}=625 m+1(m \in \mathbf{N}) .
\end{array}
$$

Also, $2^{5} \mid 2^{2000}$,
so let $2^{2000}=2^{5} \cdot n(n \in \mathbf{N})$.
From (1) and (2), we get $625 m+1=2^{5} \cdot n$.
Solving this indeterminate equation, we get
$$
\left\{\begin{array}{l}
m=32 k+15, \\
n=625 k+293
\end{array}\right.
$$
$k \in \mathbf{Z}$.
Substituting $n=625 k+293$ into (2), we get
$$
\begin{array}{l}
2^{2000}=32(625 k+293)=20000 k+9376 . \\
\therefore 2^{1999}=10000 k+4688 .
\end{array}
$$

Therefore, the last four digits of $2^{1999}$ are 4688.