For continuity at $x=5$

$\underset{x\to 5^{+}}{\mathrm{l}\mathrm{i}\mathrm{m}}f\left(x\right)=\underset{x\to 5^{-}}{\mathrm{l}\mathrm{i}\mathrm{m}}f\left(x\right)=f\left(5\right)$

$\Rightarrow \underset{x\to 5^{+}}{\mathrm{l}\mathrm{i}\mathrm{m}}\left(b\left|x-\pi \right|+3\right)=\underset{x\to 5^{-}}{\mathrm{l}\mathrm{i}\mathrm{m}}a\left(\left|\pi -x\right|+1\right)=a\left(5-\pi \right)+1$

We know that $\left|x\right|=\left\{\begin{array}{ll} x,& x\ge 0\\ -x,& x<0\end{array}\right.$

$\Rightarrow b\left(5-\pi \right)+3=a\left(5-\pi \right)+1=a\left(5-\pi \right)+1$

$\Rightarrow b\left(5-\pi \right)+3=a\left(5-\pi \right)+1$

$\Rightarrow 3-1=a\left(5-\pi \right)-b\left(5-\pi \right)$

$\Rightarrow \left(a-b\right)\left(5-\pi \right)=2$

$\Rightarrow a-b=\frac{2}{5-\pi }.$