Since a $6-$bit grayscale image is assumed, so value of pixels is $<=63$ and $>=0.($a$)$ After applying the filter, the highest result is obtained when the filter's positive values multiply $63$ and its negative values multiply $0.$The highest value that may be achieved is vmax $=(3+3+1)(63)=441.$ The minimal value will be vmin $=(-2-2-1)(63)=-315,$ according to the same logic. $($b$)$ The range of a $6-$bit picture must be $0$ to $63.$ Iout $=63($Iin vmin vmax vmin$)=63($Iin $+315756)$ where Iin and Iout are the input and output pictures, respectively, and where vmax and vmin are the maximum and minimum feasible values for