8. There are two pieces each of red, green, white, and blue (all identical except for color). Now, seven of these pieces are selected to be inlaid at the vertices of a regular hexagonal pyramid, with one piece at each vertex. The number of different inlaying methods is $\qquad$

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8. 424.

There is a color of chess pieces of which only one is selected, denoted as $a$, and there are 4 ways to select it.
(1) If $a$ is embedded at the apex of the cone, first consider the linear arrangement of the other six chess pieces, which has $\frac{6!}{2 \times 2 \times 2}=90$ ways.
The arrangement in a circle has two cases:
(i) Cyclic rings, 2 ways, each of which can be broken into 3 different linear arrangements.
(ii) Non-cyclic rings, each of which can be broken into 6 different linear arrangements. The different arrangements of the 6 chess pieces on the base are $\frac{90-2 \times 3}{6}+2=16$ ways.
In this case, there are $4 \times 16=64$ ways.
(2) If $a$ is not embedded at the apex of the cone, select one from the remaining three colored chess pieces to embed at the apex of the cone, which has 3 ways; the arrangement of the chess pieces at the base vertices has $\frac{5!}{2!\times 2!}=$ 30 ways. In this case, there are $4 \times 3 \times 30=360$ ways. In total, there are 424 ways.