jmc

Expanding the squares, we have $$\begin{array}{rl}& (3^{1001}+4^{1002}{)}^2-(3^{1001}-4^{1002}{)}^2\\ & =3^{2002}+2\cdot 3^{1001}\cdot 4^{1002}+4^{2004}\\ & -3^{2002}+2\cdot 3^{1001}\cdot 4^{1002}-4^{2004}\\ & =4\cdot 3^{1001}\cdot 4^{1002}.\end{array}$$Since $4^{1002}=4\cdot 4^{1001}$, we can rewrite the expression as $$16\cdot 3^{1001}\cdot 4^{1001}=16\cdot 12^{1001}.$$Thus, $k=\boxed{16}$.