1、Set
$$
\left\{(a, b, c, d, e) \left\lvert\, \log _{\frac{1}{6}}\left(a^{6}+b^{6}+c^{6}+d^{6}+e^{6}+1\right) \geq \log _{\frac{1}{6}} a+\log _{\frac{1}{6}} b+\log _{\frac{1}{6}} c+\log _{\frac{1}{6}} d+\log _{\frac{1}{6}} e-1\right.\right\}
$$

has a subset.
Solution: 2.
Since $a, b, c, d, e \in \mathbb{R}^{+}$
$$
\begin{array}{l} 
\log _{\frac{1}{6}}\left(a^{6}+b^{6}+c^{6}+d^{6}+e^{6}+1\right) \geq \log _{\frac{1}{6}} a+\log _{\frac{1}{6}} b+\log _{\frac{1}{6}} c+\log _{\frac{1}{6}} d+\log _{\frac{1}{6}} e-1 \\
= \log _{\frac{1}{6}} 6 a b c d e
\end{array}
$$

Therefore, $a^{6}+b^{6}+c^{6}+d^{6}+e^{6}+1 \leq 6 a b c d e$
On the other hand, by the AM-GM inequality, we have $a^{6}+b^{6}+c^{6}+d^{6}+e^{6}+1 \geq 6 a b c d e$
Thus, equality holds, giving $a=b=c=d=e=1$.
So the set is $\{(1,1,1,1,1)\}$, which has 2 subsets.