$$ { 解：f\left( x \right) =\left( x^2-x-2 \right) ^n\cos \frac{\pi}{16}x^2=\left[ \left( x-2 \right) \left( x+1 \right) \right] ^n\cos \frac{\pi}{16}x^2=\left( x-2 \right) ^n\left( x+1 \right) ^n\cos \frac{\pi}{16}x^2；} \\ { 令\left\{ \begin{array}{l} u=\left( x-2 \right) ^n\\ v=\left( x+1 \right) ^n\cos \frac{\pi}{16}x^2\\ \end{array}，则有： \right. f^{\left( n \right)}\left( x \right) \begin{array}{l} =\sum_{k=0}^n{\left( \begin{array}{c} n\\ k\\ \end{array} \right) u^{\left( k \right)}v^{\left( n-k \right)}}\\ =\sum_{k=0}^n{\left( \begin{array}{c} n\\ k\\ \end{array} \right) \left[ \left( x-2 \right) ^n \right] ^{\left( k \right)}\left[ \left( x+1 \right) ^n\cos \frac{\pi}{16}x^2 \right] ^{\left( n-k \right)}}；\\ \end{array}} \\ { f^n\left( 2 \right) =\left( \begin{array}{c} n\\ n\\ \end{array} \right) \left[ \left( x-2 \right) ^n \right] ^{\left( n \right)}\cdot \left[ \left( x+1 \right) ^n\cos \frac{\pi}{16}x^2 \right] \mid_{x=2}^{}=1\cdot n!\cdot \left( 2+1 \right) ^n\cos \left( \frac{\pi}{16}\cdot 2^2 \right) =\frac{\sqrt{2}}{2}3^nn!.} \\ { 注1：利用莱布尼茨求导公式后含有\left[ \left( x-2 \right) ^n \right] ^{\prime},\left[ \left( x-2 \right) ^n \right] '',\cdots ,\left[ \left( x-2 \right) ^n \right] ^{\left( n-1 \right)}的项代入x=2} \\ { 后，均等于0，因此这些项都不用考虑了.} \\ { 注2：\left( \begin{array}{c} n\\ k\\ \end{array} \right) 表示组合数，通常更习惯写为C_{n}^{k}.} \\ $$