\(\frac{\left(3.4.2^{16}\right)^2}{11.2^{13}.4^{11}-16^9}\)\(=\frac{3^2.4^2.2^{32}}{11.2^{12}.2^{22}-2^{36}}\)\(=\frac{3^2.2^4.2^{32}}{11.2^{34}-2^{34}.2^2}\)\(=\frac{3^2.2^{34}.2^2}{2^{34}.\left(11-2^2\right)}\)\(=\frac{2^{34}.36}{2^{34}.7}=\frac{36}{7}\)