12. If $x^{5}+3 x^{3}+1=a_{0}+a_{1}(x-1)+$ $a_{2}(x-1)^{2}+\cdots+a_{5}(x-1)^{5}$ holds for any real number $x$, then the value of $a_{3}$ is $\qquad$ (answer with a number).
12.13.
$$
\begin{array}{l}
\text { In } x^{5}+3 x^{3}+1 \\
=[(x-1)+1]^{5}+3[(x-1)+1]^{3}+1
\end{array}
$$

the coefficient of the $(x-1)^{3}$ term in the expanded form is $\mathrm{C}_{5}^{2}+3=$ 13, so, $a_{3}=13$.