\(\left| {\left| {x-3} \right|-2} \right| \le 1.\)
Неравенство вида \(\left| {f\left( x \right)} \right| \le a\) при \(a \ge 0\) равносильно двойному неравенству:
\(-a \le f\left( x \right) \le a\;\;\;\; \Leftrightarrow \;\;\;\;\,\left\{ {\begin{array}{*{20}{c}}{f\left( x \right) \le a,\;\,}\\{f\left( x \right) \ge -a.}\end{array}} \right.\)
\(\left| {\left| {x-3} \right|-2} \right| \le 1\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,-1 \le \left| {x-3} \right|-2 \le 1\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,1 \le \left| {x-3} \right| \le 3\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{\left| {x-3} \right| \le 3,}\\{\left| {x-3} \right| \ge 1}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{-3 \le x-3 \le 3,}\\{\left[ {\begin{array}{*{20}{c}}{x-3 \ge 1,\,\,}\\{x-3 \le -1}\end{array}\,\,\,\,\,\,} \right.}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{0 \le x \le 6,}\\{\left[ {\begin{array}{*{20}{c}}{x \ge 4,}\\{x \le 2}\end{array}\,\,\,\,\,\,} \right.}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{0 \le x \le 6,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{x \in \left( {-\infty ;2} \right] \cup \left[ {4; + \infty } \right)}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,x \in \left[ {0;2} \right] \cup \left[ {4;6} \right].\)
Ответ: \(\left[ {0;2} \right] \cup \left[ {4;6} \right]\).