Let probability of success is $p$ and failure is $q$
Given $n=7$,

also , $P\left(X=3\right)=5 P\left(X=4\right)$ $\Rightarrow {}^{7}C_3{p}^3{q}^4=5·{}^{7}C_4{p}^4{q}^3$

$\Rightarrow q=5p ...\left(\mathrm{i}\right)$

We know that $p+q=1\Rightarrow 6p=1$ (from $\left(\mathrm{i}\right)$)

$\therefore p=\frac{1}{6}$ and $q=\frac{5}{6}$

$\therefore$ Mean $=np=7·\frac{1}{6}=\frac{7}{6}$

and Variance $=npq=7·\frac{1}{6}·\frac{5}{6}=\frac{35}{36}$

$\therefore$ Sum of mean and variance $=\frac{7}{6}+\frac{35}{36}$ $=\frac{77}{36}$