$P\left(A\right)=\frac{2}{3},P\left(B\right)=\frac{3}{4}$
$\because A,B$ are independent events
$P\left(A\cap B\right)=P\left(A\right)·P\left(B\right)=\frac{1}{2}$
$P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)-P\left(A\cap B\right)$
$=\frac{2}{3}+\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.-\frac{1}{2}=\frac{8+9-6}{12}=\frac{11}{12}$
$P\left(\raisebox{1ex}{$A$}\!\left/ \!\raisebox{-1ex}{$A\cup B$}\right.\right)=\frac{P\left(A\cap \left(A\cup B\right)\right)}{P\left(A\cup B\right)}=\frac{P\left(A\right)}{P\left(AUB\right)}=\frac{\frac{2}{3}}{\frac{11}{12}}=\frac{8}{11}$
$P\left(\frac{A\cap B}{{\left(A\cup B\right)}^c}\right)=\frac{P\left[\left(A\cap B\right)\cap {\left(A\cup B\right)}^C\right]}{P{\left(A\cup B\right)}^C}=0$

(As $A\cap B\hspace{0.17em}\& {\left(A\cup B\right)}^C$are disjoint)
$P\left(\frac{A}{B}\right)=\frac{P\left(A\cap B\right)}{P\left(B\right)}=\frac{\frac{1}{2}}{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}=\frac{2}{3}$
$P\left(A^c\cap B^c\right)=P{\left(A\cup B\right)}^C=1-P\left(A\cup B\right)$
$=1-\frac{11}{12}=\frac{1}{12}$