Answer: -2
Analysis: Let $s=x-2$, then $2^{x}+4 x+12=0 \Leftrightarrow 2^{s}=-s-5$; Let $t=y-1$, then $\log _{2}(y-1)^{3}+3 y+12=0 \Leftrightarrow \log _{2} t=-t-5$. Note that the graphs of the functions $y=2^{x}$ and $y=\log _{2} x$ are symmetric about the line $y=x$, and the graph of the function $y=-x-5$ is also symmetric about the line $y=x$, and the x-coordinate of the intersection point of $y=x$ and $y=-x-5$ is $-\frac{5}{2}$, so $s+t=-5$, thus $x+y=s+t+3=-2$.
5. If real numbers $x, y$ satisfy $2^{x}+4 x+12=\log _{2}(y-1)^{3}+3 y+12=0$, then $x+y=$