$\text{ Length of longest interval for which }x\text{ satisfies the inequality }\left(\right|x-1|-3)\left(\right|x+2|-5)<0\text{ is }$

$\text{ We have }(\lfloor x-1\mid -3)\left(\right|x+2|-5)<0$

$$\begin{gathered} \begin{array}{r}\text{ Case I: }x\ge 1\\ \therefore (x-4)(x-3)<0\end{array} \\ \therefore x\in (3,4)-----\left(1\right) \end{gathered}$$

$\begin{array}{l}\text{ Case II: }-2\le x<1\\ \therefore (x-3)(x+2)>0\\ \therefore x\in (-\mathrm{\infty },-2)\cup (3,\mathrm{\infty })\\ \therefore x\in \varphi (\text{ as }-2\le x<1)\end{array}$

$\begin{array}{l}\text{ Case III: }x<-2\\ \therefore (x+2)(x+7)<0\\ \therefore x\in (-7,-2)------\left(2\right)\end{array}$

$\text{ From (i) and (ii), }x\in (-7,-2)\cup (3,4)$