[Solution] The edges of the small cubes with one face painted cannot coincide with the edges of the original large cube, so there are $(a-2)^{2}$ on each face, making a total of $6(a-2)^{2}$;

The small cubes with two faces painted are located along the edges of the original large cube, but the small cubes at the vertices (which have three faces painted) must be excluded. Thus, there are $a-2$ on each edge, totaling $12(a-2)$;

The small cubes with three faces painted are located at the vertices of the original large cube, totaling 8.
In summary, the answer is $(A)$.
35.9 A cube with a side length of $a \mathrm{~cm}$ (where $a$ is a natural number greater than 2) is painted red on its surface, and then cut into small cubes with a volume of $1 \mathrm{~cm}^{3}$. The number of small cubes with one, two, and three faces painted are, respectively,
(A) $6(a-2)^{2}, 12(a-2), 8$.
(B) $6(a-2)^{2}, 12(a-2), 6$.
(C) $6(a-2), 12(a-2), 8$.
(D) None of the above.