\(\begin{array}{l}x^{2}-4 x+\mathrm{m}+1=0 \\\Delta^{\prime}=2^{2}-1 \cdot(\mathrm{m}+1)=4-\mathrm{m}-1=3-\mathrm{m}\end{array}\)

Phương trình có hai nghiệm phân biệt \(\Leftrightarrow \Delta^{\prime}\gt 0 \Leftrightarrow 3-\mathrm{m}>0 \Leftrightarrow \mathrm{m}\lt 3\) (\(\left.^{*}\right)\)

Theo Vi-ét \(\left\{\begin{array}{c}x_{1}+x_{2}=4 \\ x_{1} \cdot x_{2}=\mathrm{m}+1\end{array}\right.\)

\(\begin{array}{l}x_{1}^{3}+x_{2}^{3}\lt 100 \Leftrightarrow\left(x_{1}+x_{2}\right)^{3}-3 x_{1} x_{2}\left(x_{1}+x_{2}\right)\lt 100 \\\Leftrightarrow 4^{3}-3.4 .(\mathrm{m}+1)\lt 100 \Leftrightarrow 64-12 \mathrm{~m}-12\lt 100 \Leftrightarrow-12 \mathrm{~m}\lt 48 \Leftrightarrow \mathrm{m}>-4 \text { (**) }\end{array}\)

\((*)\) và \((**)\) \(\Rightarrow-4\lt \mathrm{m}\lt 3\)

Do \(\mathrm{m} \in \mathbb{Z}\) nên \(\mathrm{m} \in\{-3 ;-2 ;-1 ; 0 ; 1 ; 2\}\)