$p(odd)=0.9*p(even)$
$=>p(1\cup 3\cup 5 )=0.9*p(2\cup 4\cup 6 )$
Again, $p(odd) + p(even)=1$
$=>0.9*p(even)+p(even)=1$
$=>1.9*p(even)=1$
$=>p(even)=1/1.9=0.5263$
$=>p(2\cup 4\cup 6)=0.5263$
Since, $p(2)=p(4)=p(6)$ as even number has the same probability
So, $p(2)=p(4)=p(6)=0.5263/3=0.175$
GIven, $p((x=4$ or $6)/x>3)=0.75$
=>$p((4\cup 6)/(4\cup 5\cup 6)=0.75$
So, =>$p(5/(4\cup 5\cup 6))=1-0.75=0.25$
Again,
$p(5)=p(5/(4∪5∪6)) *p(4∪5∪6)$
$=0.25*p(4∪5∪6) ------eq1$
$p(4∪6)=p((4∪6)/(4∪5∪6)) *p(4∪5∪6)$
$=0.75*p(4∪5∪6) -----eq2$
divide $eq1$ by $eq2$
so, $p(5)/p(4∪6)=1/3$
Since $p(4)$ and $p(6)$ are mutually exclusive that is if $4$ appears, then at same time $6$ can't appear and vice versa. So their intersection is zero.
So, $p(4∪6)=p(4)+p(6)-p(4\cap$6$)$
$=0.175+0.175+0 =0.35$
so, $p(5)=1/3 * 0.35=0.1166$
So required probability of greater than 3 is
=>$p(4)+p(5)+p(6)=0.175+0.1166+0.175=0.466$