[Solution] Given $12=2^{2} \cdot 3, 30=2 \cdot 3 \cdot 5, 42=2 \cdot 3 \cdot 7$, $\begin{array}{ll}44=2^{2} \cdot 11, & 57=3 \cdot 19, \\ 95=5 \cdot 19, & 143=11 \cdot 13,\end{array}$ $91=7 \cdot 13$, we can create the following factor table:
\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline$y$ & 2 & 3 & 5 & 7 & 11 & 13 & 19 \\
\hline 12 & $\vee v$ & $\checkmark$ & & & & & \\
\hline 30 & $\checkmark$ & $\checkmark$ & $v$ & & & & \\
\hline 42 & $\vee$ & $v$ & & $\checkmark$ & & & \\
\hline 44 & $\vee v$ & & & & $v$ & & \\
\hline 57 & & $\checkmark$ & & & & & $v$ \\
\hline 91 & & & & $\checkmark$ & & $\vee$ & \\
\hline 95 & & & $v$ & & & & $v$ \\
\hline 143 & & & & & $v$ & $v$ & \\
\hline
\end{tabular}

From the table, we can see: For groups with the same product, 95 and 57 should be in different groups. Similarly, 91 and 143; 44 and 143; 42 and 91; 30 and 95, etc., should all be in different groups. This way, we can divide them into two groups: 12, 42, 95, 143 and 30, 44, 57, 91.
Therefore, the answer is $(C)$.
$28 \cdot 30$ Divide the 8 numbers $12,30,42,44,57,91,95,143$ into two groups so that the product of the numbers in each group is equal. The correct grouping is
(A) $12,42,57,143$ and $30,44,91,95$.
(B) $12,30,95,143$ and $42,44,57,91$.
(C) $12,42,95,143$ and $30,44,57,91$.
(D) $12,44,95,143$ and $30,42,57,91$.
(3rd "Five Sheep Cup" Junior High School Mathematics Competition, 1991)