a) Let a,b $\epsilon$ (G_{1} $\cap$ G_{2})
Now, a,b $\epsilon$ G_{1} $\Rightarrow$ a*b $\epsilon$ G_{1 }($\because$ G_{1} is algebraic structure)
Similarlly a^{-1}, b^{-1} $\epsilon$ G_{1} ($\because$ G_{1} is group)
Also this will be true for G_{2}.
$\therefore$ As a,b $\epsilon$ (G_{1} $\cap$ G_{2}), a^{-1}, b^{-1} $\epsilon$ (G_{1} $\cap$ G_{2}), which will make it subgroup of G.
b) G_{1 }$\cup$ G_{2 }need not be a subgroup.
For example, Let G = {1,3,5,7} is a group w.r.t multiplication modulo 8.
Consider G_{1} = {1,3} and G_{2} = {1,5}
Here G_{1 }& G_{2 }are subgroups of G but G_{1 }$\cup$ G_{2} is not a subgroup of G.