Let E be the ellipse (x^2/(16)+(y^2/(7)=1 and let H be the hyperbola having the same foci as E and eccentricity 3/2.

For $E$: $a^2=16$, $b^2=7$, so $c^2=9$ and the foci are $(\pm3,0)$. For $H$ the foci are the same and $e=\frac32$, so $a_H=\frac{c}{e}=2$ and $b_H^2=9-4=5$, giving $\frac{x^2}{4}-\frac{y^2}{5}=1$: (A) true. Solving the two equations gives $x^2=\frac{64}{9}$, $y^2=\frac{35}{9}$, hence four points $\left(\pm\frac83,\pm\frac{\sqrt{35}}{3}\right)$: (B) true. The slopes of the tangents at a common point are $-\frac{7x}{16y}$ and $\frac{5x}{4y}$, whose product is $-\frac{35x^2}{64y^2}=-1$ at these points, so they are perpendicular (confocal conics cut orthogonally): (C) true. The rectangle has sides $\frac{16}{3}$ and $\frac{2\sqrt{35}}{3}$ and area $\frac{32\sqrt{35}}{9}\approx21.04\ne20$: (D) false.