(1)

Let the parent nucleus be $X$ which breaks into two fragments $Y$

${}^{240}X\to 2^{120}Y$

Binding energy of $X$, ${\text{B.E.}}_X=240\times 7.6=1824$ MeV

Binding energy of $Y$, ${\text{B.E.}}_Y=120\times 8.5=1020$ MeV

Energy released, $E=2{\left(\text{B.E.}\right)}_Y-{\left(\text{B.E.}\right)}_X$

           $=2\times \left(1020\right)-1824=216$ $\text{MeV}$  

(3)

Given: energy, $E={10}^{-20}\text{J}$

Now, $1 \text{J}=\frac{1}{1.6\times {10}^{-19}}$ $\text{eV}$

$\therefore$   $E=\frac{{10}^{-20}}{1.6\times {10}^{-19}} \text{eV}=0.0625 \text{eV}\simeq 0.06 \text{eV}$

(2)

Given mass $m=0.5$$\text{g}=0.5\times {10}^{-3}\mathrm{kg}$

According to Einstein mass-energy equivalence,

       $E=m{c}^2=0.5\times {10}^{-3}\times {(3\times {10}^8)}^2=4.5\times {10}^{13}\text{J}$