The value of equilibrium constant of the reaction $HI(g)\rightleftharpoons \frac{1}{2}{H}_2(g)+\frac{1}{2}{I}_2(g)$ is $8.0$. The equilibrium constant of the reaction $H_2(g)+I_2(g)\rightleftharpoons 2HI(g)$ will be:

Let the first reaction be (1): $HI\rightleftharpoons \frac{1}{2}{H}_2+\frac{1}{2}{I}_2$ with $K_1=8$.Reverse of reaction (1) is: $\frac{1}{2}{H}_2+\frac{1}{2}{I}_2\rightleftharpoons HI$ with $K^{\prime }=\frac{1}{K_1}=\frac{1}{8}$.The target reaction is $H_2+I_2\rightleftharpoons 2HI$, which is $2\times$ the reversed reaction.When a reaction is multiplied by a factor $n$, the new equilibrium constant $K_{new}=(K^{\prime }{)}^n$.
$$K_{new}={\left(\frac{1}{8}\right)}^2=\frac{1}{64}$$