#### Answer

(a) The net gravitational force on mass A due to masses B and C is $2.77\times 10^{-8}~N$ directed to the right.
(b) The net gravitational force on mass A due to masses B and C is $2.50\times 10^{-8}~N$ directed to the left.

#### Work Step by Step

Let a force directed to the right be positive.
(a) We can find the net gravitational force on mass A due to masses B and C.
$F = \frac{GM_B~M_A}{R_{AB}^2}+ \frac{GM_C~M_A}{R_{AC}^2}$
$F = \frac{(6.67\times 10^{-11}~m^3/kg~s^2)(2.00~kg)(2.00~kg)}{(0.50~m)^2}+ \frac{(6.67\times 10^{-11}~m^3/kg~s^2)(2.00~kg)(2.00~kg)}{(0.10~m)^2}$
$F = 2.77\times 10^{-8}~N$
The net gravitational force on mass A due to masses B and C is $2.77\times 10^{-8}~N$ directed to the right.
(b) We can find the net gravitational force on mass A due to masses B and C.
$F = \frac{GM_B~M_A}{R_{AB}^2}- \frac{GM_C~M_A}{R_{AC}^2}$
$F = \frac{(6.67\times 10^{-11}~m^3/kg~s^2)(2.00~kg)(2.00~kg)}{(0.40~m)^2}- \frac{(6.67\times 10^{-11}~m^3/kg~s^2)(2.00~kg)(2.00~kg)}{(0.10~m)^2}$
$F = -2.50\times 10^{-8}~N$
The net gravitational force on mass A due to masses B and C is $2.50\times 10^{-8}~N$ directed to the left.