\lim_{x\to 0}\left(\frac{\cos 3x}{\cos 4x}\right)^{\frac 1 {x^2}}=\lim_{x\to 0}e^{\frac 1 {x^2}\ln\left(\frac{\cos 3x}{\cos 4x}\right)}=e^{\displaystyle\lim_{x\to 0}\frac{\ln(\cos 3x)-\ln(\cos 4x)}{x^2}}=e^{\displaystyle\lim_{x\to 0}\frac{\frac{3}{\cos 3x}(-\sin 3x)-\frac{4}{\cos 4x}(-\sin 4x)}{2x}}=e^{\displaystyle\lim_{x\to 0}\frac{4\hbox{tg}\,4x-3\hbox{tg}\,3x}{2x}}=e^{\displaystyle\lim_{x\to 0}\frac{\frac{16}{\cos^24x}-\frac{9}{\cos^23x}}2}=e^{\frac 7 2}