Let $f:(0,\infty )\to R$ and $F\left(x\right)=\underset{0}{\overset{x}{\int }}f\left(t\right)dt$. If $F\left(x^2\right)=x^2(1+x)$ then $f\left(4\right)$ equals

Let a function $f:\left[0,5\right]\to R$ be continuous, $f\left(1\right)=3$ and $F$ be defined as: $F\left(x\right)={\int }_1^x{t}^{2}g\left(t\right)dt,$ where $g\left(t\right)={\int }_1^{t}f\left(u\right)du.$ Then for the function $F\left(x\right),$ the point $x=1$ is:

The least value of the function $f\left(x\right)={\int }_0^x\left(3\sin x+4\cos x\right)dx$ in the interval $\left[\frac{5\pi }{4},\frac{4\pi }{3}\right]$ is