= Solution
For positive operators, the unique <positive square root of an operator> respects tensor products: $\sqrt{A\otimes B}=\sqrt A\otimes\sqrt B$. To verify this, diagonalize $A,B$; the product eigenvectors have eigenvalues $a_ib_j$, whose square roots are $\sqrt{a_i}\sqrt{b_j}$. Hence the operator inside the outer square root in the fidelity factors as
$$
\bigl(\sqrt\rho\,\rho'\sqrt\rho\bigr)\otimes
\bigl(\sqrt\sigma\,\sigma'\sqrt\sigma\bigr).
$$
Taking its positive square root and using $\operatorname{Tr}(C\otimes D)=\operatorname{Tr}C\,\operatorname{Tr}D$ gives \b[$\boxed{F(\rho\otimes\sigma,\rho'\otimes\sigma')=F(\rho,\rho')F(\sigma,\sigma')}$.]
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