Let $$f:R \to R$$ be a continuous odd function, which vanishes exactly at one point and $$f\left( 1 \right) = {1 \over {2.}}$$ Suppose that $$F\left( x \right) = \int\limits_{ - 1}^x {f\left( t \right)dt} $$ for all $$x \in \,\,\left[ { - 1,2} \right]$$ and $$G(x)=$$ $$\int\limits_{ - 1}^x {t\left| {f\left( {f\left( t \right)} \right)} \right|} dt$$ for all $$x \in \,\,\left[ { - 1,2} \right].$$ If $$\mathop {\lim }\limits_{x \to 1} {{F\left( x \right)} \over {G\left( x \right)}} = {1 \over {14}},$$ then the value of $$f\left( {{1 \over 2}} \right)$$ is

Let $$F\left( x \right) = \int\limits_x^{{x^2} + {\pi \over 6}} {2{{\cos }^2}t\left( {dt} \right)} $$ for all $$x \in R$$ and $$f:\left[ {0,{1 \over 2}} \right] \to \left[ {0,\infty } \right]$$ be a continuous function. For $$a \in \left[ {0,{1 \over 2}} \right],\,$$ $$F'(a)+2$$ is the area of the region bounded by $$x=0, y=0, y=f(x)$$ and $$x=a,$$ then $$f(0)$$ is