$$=, 1.6 \text {. }$$

Given $x=\frac{3-\sqrt{5}}{2}$, we know that
$$\begin{array}{l} x^{2}-3 x+1=0, x+\frac{1}{x}=3 . \\ \text { Therefore, } x^{3}-3 x^{2}+3 x+\frac{6}{x^{2}+1} \\ =x\left(x^{2}-3 x\right)+3 x+\frac{6}{x^{2}+1}=2 x+\frac{6}{3 x} \\ =2 x+\frac{2}{x}=2\left(x+\frac{1}{x}\right)=6 . \end{array}$$