#### Answer

(a) The scale reading is $0.60~W$
(b) The scale reading is $0.64~W$

#### Work Step by Step

(a) Let $\rho_w$ be the density of water. Then the stone has a density of $2.50~\rho_w$. Let $V$ be the volume of the stone. We can write an expression for the weight of the stone:
$W = 2.50~\rho_w~V~g$
According to Archimedes' principle, the buoyant force is equal to the weight of the water that is displaced. We can find the buoyant force:
$F_b = \rho_w~V~g = \frac{W}{2.50} = 0.40~W$
We can consider the forces on the stone to find the scale reading $F_s$:
$\sum F = 0$
$F_s+F_b-W = 0$
$F_s = W - F_b$
$F_s = W - 0.40~W$
$F_s = 0.60~W$
The scale reading is $0.60~W$
(b) According to Archimedes' principle, the buoyant force is equal to the weight of the oil that is displaced. We can find the buoyant force:
$F_b = \rho_o~V~g$
$F_b = 0.90~\rho_w~V~g$
$F_b = 0.90~(\frac{W}{2.50})$
$F_b = 0.36~W$
We can consider the forces on the stone to find the scale reading $F_s$:
$\sum F = 0$
$F_s+F_b-W = 0$
$F_s = W - F_b$
$F_s = W - 0.36~W$
$F_s = 0.64~W$
The scale reading is $0.64~W$.