\( \begin{array}{l}\left(1-\frac{1}{2} x\right)^{5}=C_{5}^{0}+C_{5}^{1}\left(-\frac{1}{2} x\right)+C_{5}^{2}\left(-\frac{1}{2} x\right)^{2}+C_{5}^{3}\left(-\frac{1}{2} x\right)^{3}+C_{5}^{4}\left(-\frac{1}{2} x\right)^{4}+C_{5}^{5}\left(-\frac{1}{2} x\right)^{5} \\=1-\frac{5}{2} x+\frac{5}{2} x^{2}-\frac{5}{4} x^{3}+\frac{5}{16} x^{4}-\frac{1}{32} x^{5}=a_{0}+a_{1} x+a_{2} x^{2}+a_{3} x^{3}+a_{4} x^{4}+a_{5} x^{5}\end{array}\)

Suy ra: \(a_{0}=1, a_{1}=-\frac{5}{2}, a_{2}=\frac{5}{2}, a_{3}=-\frac{5}{4}, a_{4}=\frac{5}{16}, a_{5}=-\frac{1}{32}\).

Ta thấy hệ số lớn nhất tìm được là \(a_{2}=\frac{5}{2}\).

Thay \(x=1\) vào \(\left(^{*}\right)\), ta được: \(\left(1-\frac{1}{2}\right)^{5}=a_{0}+a_{1}+a_{2}+a_{3}+a_{4}+a_{5}\).

Vậy \(a_{0}+a_{1}+a_{2}+a_{3}+a_{4}+a_{5}=\frac{1}{32}\).