Conservation of Momentum

Masses of fragments: $\frac{2m}{5}, \frac{2m}{5}, \frac{m}{5}$ By conservation of momentum (initial momentum = 0): $0 = \frac{2m}{5}(-v\hat{i}) + \frac{2m}{5}(-v\hat{j}) + \frac{m}{5}\vec{v}'$ $\Rightarrow \vec{v}' = 2v\hat{i} + 2v\hat{j}$ $v' = \sqrt{(2v)^2 + (2v)^2} = 2\sqrt{2}\,v$

By conservation of momentum, the third part balances the resultant of the first two:
$m(4) = \sqrt{(1\times12)^2 + (2\times8)^2} = 20$
$m = 5$ kg

The bomb is at rest, so its initial momentum is zero. By conservation of momentum the two pieces have equal and opposite momenta.
Momentum of the 18 kg piece $= 18 \times 6 = 108$ kg m/s
So the 12 kg piece also has momentum $p = 108$ kg m/s (in the opposite direction).
Kinetic energy: $KE = \dfrac{p^2}{2m}$
$KE = \dfrac{(18 \times 6)^2}{2 \times 12} = \dfrac{11664}{24}$
$KE = 486$ J
(The velocity of the 12 kg piece is $\dfrac{108}{12} = 9$ m/s, and $\dfrac{1}{2} \times 12 \times 9^2 = 486$ J.)