It's probably simplest to do (a) by using "conservation of energy". Taking the table top as 0 potential energy, the initial potential energy of the chain is the integral, from 0 to l, of (mg)/l x dx which is (1/2)mgl (the potential energy if the chain had all been at its midpoint- that's reasonable!). If, at given time t, the upper end of the chain has fallen fallen a distance y, then all but l-y is now on the table and so has 0 potential energy. The remaining, still falling, part of the chain has potential energy (1/2)(mg/l)(l-y)^2 by the same integration. The kinetic energy of the chain must be the difference, (1/2)(mg/l)(2yl- y^2) and that must be equal to 1/2 mv^2= 1/2 m (dy/dt)^2. y must satisfy the differential equation dy/dt= sqrt((mg/l)(2yl- y^2). And, of course, dy/dt is the speed asked for in part (a).