(a) Your calculation is very good but you have not reached the end of it.
The distance of $2.5m$ which the disk travels across the belt is measured diagonally. It is the hypotenuse of a $3:4:5$ triangle. The perpendicular distance across the belt is $2.0m$. So that is the minimum width.
(b) This is the same problem as in A block is pushed onto a conveyor belt.
Initially the velocity of the disk over the ground is $OA=4m/s$ along the $Oy$ axis. When the disk has come to rest relative to the conveyor belt its velocity over the ground is $OB=3m/s$ along the $Ox$ axis. At intermediate times the velocity of the disk over the ground is represented by points on the line $AB$.
At point $C$ the magnitude $OC$ of the velocity over the ground is minimum, where $OC\perp AB$. $OAB$ and $COB$ are similar triangles. Therefore $OC=\frac45 \times 3=2.4m/s$.