Answer
(a) The speed and direction of Piece 1 is $8.00\times 10^3~m/s$ away from the Earth.
The speed and direction of Piece 2 is $5.20\times 10^3~m/s$ away from the Earth.
(b) The explosion supplied $7.0 \times 10^8~J$ of energy.
Work Step by Step
(a) Let $v$ be the original speed. Let $v_r$ be the relative speed of the two pieces.
Since the two parts have equal mass, then one part must move with a speed of $v+v_r/2$ and the other piece must move with a speed of $v-v_r/2$. Then the original momentum is conserved, and the relative speed between the two pieces is $v_r$.
Piece 1:
$v+v_r/2 = 6.60\times 10^3~m/s + 1.40\times 10^3~m/s$
$v+v_r/2 = 8.00\times 10^3~m/s$
The speed and direction of Piece 1 is $8.00\times 10^3~m/s$ away from the Earth.
Piece 2:
$v-v_r/2 = 6.60\times 10^3~m/s - 1.40\times 10^3~m/s$
$v-v_r/2 = 5.20\times 10^3~m/s$
The speed and direction of Piece 2 is $5.20\times 10^3~m/s$ away from the Earth.
(b) Before the explosion:
$KE_0 = \frac{1}{2}(725~kg)(6.60\times 10^3~m/s)^2$
$KE_0 = 1.58\times 10^{10}~J$
After the explosion:
$KE = \frac{1}{2}(725~kg/2)(8.00\times 10^3~m/s)^2 + \frac{1}{2}(725~kg/2)(5.20\times 10^3~m/s)^2$
$KE = 1.65\times 10^{10}~J$
$KE-KE_0 = 1.65\times 10^{10}~J - 1.58\times 10^{10}~J$
$KE-KE_0 = 7.0 \times 10^8~J$
The explosion supplied $7.0 \times 10^8~J$ of energy.