$$f\left( {{a}_{1}} \right)+f\left( {{a}_{2}} \right)+\cdots +f\left( {{a}_{5}} \right)=\sum\limits_{i=1}^{5}{\left( 2{{a}_{i}}-\cos {{a}_{i}} \right)}$$  $$=10{{a}_{3}}-\left[ \cos \left( {{a}_{3}}-\frac{ \pi }{4} \right)+\cos \left( {{a}_{3}}+\frac{ \pi }{4} \right) \right]-\left[ \cos \left( {{a}_{3}}-\frac{ \pi }{8} \right)+\cos \left( {{a}_{3}}+\frac{ \pi }{8} \right) \right]-\cos {{a}_{3}}$$  $$=10{{a}_{3}}-2\cos {{a}_{3}}\cos \frac{ \pi }{4}-2\cos {{a}_{3}}\cos \frac{ \pi }{8}-\cos {{a}_{3}}$$  令$$g\left( {{a}_{3}} \right)=10{{a}_{3}}-\left( \sqrt{2}+2\cos \frac{ \pi }{8}+1 \right)\cos {{a}_{3}}$$，则$${g}'\left( {{a}_{3}} \right)=10+\left( \sqrt{2}+2\cos \frac{ \pi }{8}+1 \right)\sin {{a}_{3}}\textgreater0$$，$$g\left( {{a}_{3}} \right)$$在$$\mathbf{R}$$上单调增，又$$g\left( \frac{ \pi }{2} \right)=5 \pi $$，所以$${{a}_{3}}=\frac{ \pi }{2}$$．  从而，$${{a}_{1}}=\frac{ \pi }{4}{{a}_{5}}=\frac{3}{4} \pi $$，$${{\left[ f\left( {{a}_{3}} \right) \right]}^{2}}-{{a}_{1}}{{a}_{5}}={{\left( ~\pi -\cos \frac{ \pi }{2} \right)}^{2}}-\frac{ \pi }{4}\cdot \frac{3}{4} \pi =\frac{13}{16}{{ \pi }^{2}}$$．  故选$$\text{D}$$． 
设函数$$f\left( x \right)=2x-\cos x$$，$$\left { {{a}_{n}} \right }$$是公差为$$\frac{ \pi }{8}$$的等差数列，$$f\left( {{a}_{1}} \right)+f\left( {{a}_{2}} \right)+\cdots +f\left( {{a}_{5}} \right)=5 \pi $$，则$${{\left[ f\left( {{a}_{3}} \right) \right]}^{2}}-{{a}_{1}}{{a}_{5}}=$$（~ ~ ）．

[A] $$0$$ 
[B] $$\frac{1}{16} \pi $$ 
[C] $$\frac{1}{8} \pi $$ 
[D] $$\frac{13}{16}{{ \pi }^{2}}$$ 