a) Let a,b $\epsilon$ (G1 $\cap$ G2)
Now, a,b $\epsilon$ G1 $\Rightarrow$ a*b $\epsilon$ G1 ($\because$ G1 is algebraic structure)
Similarlly a-1, b-1 $\epsilon$ G1 ($\because$ G1 is group)
Also this will be true for G2.
$\therefore$ As a,b $\epsilon$ (G1 $\cap$ G2), a-1, b-1 $\epsilon$ (G1 $\cap$ G2), which will make it subgroup of G.
b) G1 $\cup$ G2 need not be a subgroup.
For example, Let G = {1,3,5,7} is a group w.r.t multiplication modulo 8.
Consider G1 = {1,3} and G2 = {1,5}
Here G1 & G2 are subgroups of G but G1 $\cup$ G2 is not a subgroup of G.