Given $f\left(x\right)=\left\{\begin{array}{ll}{\int }_0^x\left(5-\left|t-3\right|\right)dt,& x>4\\ x^2+bx,& x\le 4\end{array}\right.$

Also given, $f\left(x\right)$ is continuous at $x=4$

So $\underset{x\to 4^{-}}{\lim }f\left(x\right)=\underset{x\to 4^{+}}{\lim }f\left(x\right)=f\left(4\right)$

So $16+4b={\int }_0^3\left(2-t\right)dt+{\int }_3^4\left(8-t\right)dt$

$\Rightarrow 16+4b={\left(2t-\frac{t^2}{2}\right)}_0^3+{\left(8t-\frac{t^2}{2}\right)}_3^4$

$\Rightarrow 16+4b=15$

So $b=\frac{-1}{4}$

Now differentiating $f\left(x\right)$ we get,

$f^{\text{'}}\left(x\right)=\left\{\begin{array}{ll}\left(5-\left|x-3\right|\right),& x>4\\ 2x+b,& x<4\end{array}\right.$

Now at $x=4$

$\mathrm{LHD}=2x+b=2\times 4-\frac{1}{4}=\frac{31}{4}$

$\mathrm{RHD}=5-\left|x-3\right|=4$

$\mathrm{LHD}\ne \mathrm{RHD}$

So, option $\left(A\right)$ is true

And $f^{\text{'}}\left(3\right)+f^{\text{'}}\left(5\right)=\frac{23}{4}+3=\frac{35}{4}$

So, option $\left(B\right)$ is true

Now $f\left(x\right)=x^2-\frac{x}{4}$ at $x\le 4$

$f^{\text{'}}\left(x\right)=2x-\frac{1}{4}$

Thus the function is not increasing in the interval in $x\in \left(-\infty ,\frac{1}{8}\right)$

So, option $\left(C\right)$ is NOT TRUE.

And function $f\left(x\right)$ is also have local minima at $x=\frac{1}{8}$ so option $\left(D\right)$ is also true.