[Solution] (1) The proposition is not true when $a=-b$;
(2) From the given, we know $a b>0$, and $a b^{4}a b^{2}$. Therefore, we have $a b>a b^{4}>a b^{2}$;
(3) If the diagonals of quadrilateral $A B C D$ intersect at $O$, and divide the quadrilateral into four triangles of equal area.
$$
\begin{array}{ll}
\because & S_{\triangle A D O}=S_{\triangle D C O}, \quad \therefore \quad A O=O C ; \\
\text { Also } S_{\triangle C D O}=S_{\triangle C B O}, \quad \therefore \quad D O=O B,
\end{array}
$$

Since the diagonals bisect each other, the quadrilateral is a parallelogram. Therefore, proposition (3) is true.
For (4), from the given, we know $E G=2 B C, H F=2 A B$ (note that $E, P, G$ and $F, P, H$ are collinear).
$$
\begin{aligned}
S_{E F C H} & =S_{\triangle E F G}+S_{\triangle E H T} \\
& =\frac{1}{2} \cdot E G \cdot F P+\frac{1}{2} \cdot E G \cdot H P \\
& =\frac{1}{2} \cdot E G(F P+H P) \\
& =\frac{1}{2} \cdot E G \cdot H F=2 \cdot B C \cdot A B \\
& =2 S_{\text {E }} \cdot
\end{aligned}
$$

Thus, the conclusion is correct.
In summary, there are 3 correct propositions.
Therefore, the answer is $(C)$.
$18 \cdot 20$ Among the following four propositions:
(1) If $a, b$ are real numbers, then $(a+b)^{0}=1$;
(2) If $a a b^{4}>a b^{2}$;
(3) If a quadrilateral is divided into four triangles of equal area by its two diagonals, then this quadrilateral is a parallelogram;
(4) If $P$ is any point inside rectangle $A B C D$, and points $E, F, G$, $H$ are the reflections of point $P$ about the four sides of the rectangle, then $S_{E F Y H}$ $=2 S_{A B C D}$.
The number of correct propositions is
(A) 1 .
(B) 2 .
(C) 3 .
(D) 4 .
(China Sichuan Junior High School Mathematics Competition, 1992)