According to the given condition, we have random variables

$11, 22, 33 \mathrm{and} 44$

As, $P\left(X=r\right)=K{r}^3$ where, $r=1, 2, 3, 4$

Hence, 

$P\left(X=1\right)=K{\left(1\right)}^3=K$,

$P\left(X=2\right)=K{\left(2\right)}^3=8K$,
$P\left(X=3\right)=K{\left(3\right)}^3=27K$ and 

$P\left(X=4\right)=K{\left(4\right)}^3=64K$
As, we know, here

$\sum _1^{4}P\left(X=r\right)=1$    (sum of probabilities)

$\Rightarrow K+8K+27K+64K=1 \Rightarrow K=\frac{1}{100}$

Now,

$P\left(\frac{1}{2}<X<\frac{5}{2}\mid X>1\right)=\frac{P\left[\left\{1, 2\right\}\cap \left\{ 2, 3, 4\right\}\right]}{P\left\{ 2, 3, 4\right\}}=\frac{P\left(X=2\right)}{P\left(X=2\right)+P\left(X=3\right)+P\left(X=4\right)}$

$=\frac{P\left(X=2\right)}{P\left(X=2\right)+P\left(X=3\right)+P\left(X=4\right)}=\frac{8K}{8K+27K+64K}=\frac{8}{99}$