$2P\left(X=1\right)=3P\left(X=2\right)=P\left(X=3\right)=5P\left(X=4\right)$

Let $P\left(X=3\right)=\alpha$.

$\sum P\left(X\right)=1$

$\Rightarrow P\left(1\right)+P\left(2\right)+P\left(3\right)+P\left(4\right)=1$

$\Rightarrow \frac{\alpha }{2}+\frac{\alpha }{3}+\alpha +\frac{\alpha }{5}=1$

$\Rightarrow \frac{61}{30}\alpha =1$

$\Rightarrow \alpha =\frac{30}{61}$

Similarly, all values can be found out.

Mean $\left(\mu \right)=E\left(X\right)=\sum _{X=1}^{4}X{P}_X\left(X\right)$

$=\left(1\right)\left(\frac{15}{61}\right)+\left(2\right)\left(\frac{10}{61}\right)+\left(3\right)\left(\frac{30}{61}\right)+\left(4\right)\left(\frac{6}{61}\right)$

$\therefore \mu =\frac{149}{61}$

$\Rightarrow {\left(\mu \right)}^2=5.9666$

Variance $\left({\sigma }^2\right)=\sum ^4{\left(x-\mu \right)}^2{P}_x\left(x\right)$

$={\left(x-\frac{149}{61}\right)}^2\left(\frac{15}{6}\right)+{\left(x-\frac{149}{61}\right)}^2\left(\frac{10}{61}\right)+{\left(x-\frac{149}{61}\right)}^2\left(\frac{30}{61}\right)+{\left(x-\frac{149}{61}\right)}^2\left(\frac{6}{61}\right)$

$\therefore {\sigma }^2=0.935224$

${\mu }^2+{\sigma }^2=5.9666+0.935224$

$=6.901224$

${\mu }^2+{\sigma }^2=\frac{421}{61}$.