$P\left(\{1,3,5\}\right) = 0.9 P\left(\{2,4,6\}\right)$ and their sum must be $1$. So,
$P\left(\{1,3,5\}\right) = \frac{0.9}{1.9} = 0.4736$ and
$P\left(\{2,4,6\}\right) = \frac{1}{1.9} = 0.5263$
Given that probability of getting $2$ or $4$ or $6$ is same.
So, $P(2) = P(4) = P(6) = \frac{0.5263}{3} = 0.1754$
$\quad\; P\left(\{4,6\}\right) \mid x > 3)= 0.75$
$\Rightarrow P(5 \mid x>3) = 0.25$
$\Rightarrow P(5) = \frac{1}{3} (P(4)+ P(6)) \because x> 3$
$\Rightarrow x \in \{4,5,6\}$
$\Rightarrow P(5) = \frac{2.P(4)}{3} \because P(4) = P(6)$.
So, $P(x > 3) = P(4) + P(5) + P(6) = \frac{8}{3}\times 0.1754 =0.468 $