3. $6^{11}+C_{11}^{1} 6^{10}+C_{11}^{2} 6^{9}+\cdots+C_{11}^{10} 6-1$ when divided by 8 yields a remainder of $\qquad$ .
3.5.

Notice that,
$$
\begin{array}{l}
6^{11}+\mathrm{C}_{11}^{1} 6^{10}+\mathrm{C}_{11}^{2} 6^{9}+\cdots+\mathrm{C}_{11}^{10} 6-1 \\
=\mathrm{C}_{10}^{0} 6^{11}+\mathrm{C}_{11}^{1} 6^{10}+\cdots+\mathrm{C}_{11}^{10} 6+\mathrm{C}_{11}^{11} 6^{0}-2 \\
=(6+1)^{11}-2=7^{11}-2 \\
\equiv(-1)^{11}-2 \equiv 5(\bmod 8) .
\end{array}
$$

Therefore, the required result is 5.