12. Given the equation $x^{5}-x^{2}+5=0$ with its five roots being $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}, f(x)=x^{2}+1$, then $\prod_{k=1}^{5} f\left(x_{k}\right)=$
Since $x^{5}-x^{2}+5=\left(x-x_{1}\right)\left(x-x_{2}\right)\left(x-x_{3}\right)\left(x-x_{4}\right)\left(x-x_{5}\right)$,
let $x=\mathrm{i}$, we get $\mathrm{i}+6=\left(\mathrm{i}-x_{1}\right)\left(\mathrm{i}-x_{2}\right)\left(\mathrm{i}-x_{3}\right)\left(\mathrm{i}-x_{4}\right)\left(\mathrm{i}-x_{5}\right)$,
let $x=-\mathrm{i}$, we get $-\mathrm{i}+6=\left(-\mathrm{i}-x_{1}\right)\left(-\mathrm{i}-x_{2}\right)\left(-\mathrm{i}-x_{3}\right)\left(-\mathrm{i}-x_{4}\right)\left(-\mathrm{i}-x_{5}\right)$, multiplying the above two equations, we get $\prod_{k=1}^{5} f\left(x_{k}\right)=(6+\mathrm{i})(6-\mathrm{i})=37$.