HINT $\rm \ Notice\ that\ \ mod\ a\: m:\ \ a\: b\ \equiv\ \ a\ (b\ mod\ m),\ $ since $\rm\ \ a\: m\ |\ a\: (b - b\ mod\ m)\:.\ $
$\rm So\ \ \ mod\ 16\cdot 25:\ \ 2^{400} = 16\cdot 2^{396}\ \equiv\ 16\ (2^{396}\: mod\ 25)\:.\: $ By $\rm\ \phi(p^2) = p\:(p-1)\ $ and little Euler
$\rm \phi(25) = 20\ \ \Rightarrow\ \ mod\ 25:\ \ 2^{20}\: \equiv\ 1\ \Rightarrow\ 2^{396}\ \equiv \frac{(2^{20})^{20}}{4^2}\ \equiv\ \left(\frac{1}{4}\right)^2\ \equiv\ (-6)^2\ \equiv\ 11\quad $
Thus $\rm\ a\:b\ \equiv\ a\: (b\ mod\ m)\ \equiv\ 16\cdot 11 \pmod{16\cdot 25}\:.$
NOTE $\ $ The congruence mentioned in the first line above frequently proves handy for congruence arithmetic, so it is well worth committing to memory. As remarked, one could alternatively employ $\rm\ 2^{10}\: =\ 1024\ \equiv\: -1\pmod{25}\:,\:$ but, as the proverb says, if you give a student one $\phi$ value then you feed them one answer, but teach a student how to $\phi$ and you feed them answers for a lifetime.