$\begin{array}{l}\text { 2. If } x^{3}+5 x^{2}-7 x-3 \\ =(x-4)^{3}+a(x-4)^{2}+b(x-4)+c \text {, } \\ \text { then }(a, b, c)=\end{array}$
2. $(17,81,113)$.

Notice that
$$
\begin{array}{l}
(x-4)^{3}+a(x-4)^{2}+b(x-4)+c \\
=x^{3}+(a-12) x^{2}+(b-8 a+48) x+ \\
\quad(16 a-4 b+c-64) .
\end{array}
$$

From $a-12=5, b-8 a+48=-7$,
$$
16 a-4 b+c-64=-3 \text {, }
$$

Solving these, we get $a=17, b=81, c=113$.
Therefore, $(a, b, c)=(17,81,113)$.