解：?$(1)$?解方程?$4x-(x+5)=1$?得?$x=2$?
解方程?$-2y-y=3$?得?$y=-1$?
∵?$2+(-1)=1$?，滿足方程的解之和為?$1$?
∴方程?$4x-(x+5)=1$?與方程?$-2y-y=3$?是?$“$?美好方程?$”$?
?$(2)$?解方程?$3x-2=x+4$?，得?$x=3$?
若兩個(gè)方程是“美好方程”
則方程?$\frac {x}2+m=0$?的解為?$1-3=-2$?
將?$x=-2$?代入方程可得?$\frac {-2}2+m=0$?，?$m=1$?
?$(3)$?解方程?$\frac {1}{2024}x-1=0$?，得?$x=2024$?
?$\frac {1}{2024}(y+2)+1=3y+k+6$?可化為?$\frac {1}{2024}(y+2)+1=3(y+2)+k$?
則由題可知關(guān)于?$x $?的方程?$\frac {1}{2024}x-1=0$?與關(guān)于?$(y+2)$?的
方程?$\frac {1}{2024}(y+2)+1=3(y+2)+k$?是?$“$?美好方程?$”$?
∴方程?$\frac {1}{2024}(y+2)+1=3(y+2)+k$?的解為?$1-2024=-2023$?
即?$y+2=-2023$?，?$y=-2025$?
