differential question help

Can someone please explain Why do they choose 2 as the value that v approaches for the last part of the question?

You have the differential equation
1/2 dv/dt = 2 - v
As v -> 2, the right hand side -> 0 so the derivative -> 0, so constant value.

Im presuming part b) had something like
v ~ e^(-2t)
so the "-v" on the right hand side of the differential equation corresponds to a decaying exponential. Thats why in this part you kow that v will converge to a constant value and that corresponds to the right hand side being 0 so v=2. If it had been "+v", then it would have represented an exponentially increasinng term, rather than a decaying exponential and v=2 would have been an unstable steady state, so v would move away from 2 if it was perturbed slightly.