Work in a Spring

don't neglect gravity

It's a bit tricky to understand.

When you use .5kx^2 with x measured from the equilibrium position, you are really finding the total potential energy change of the system including both spring PE plus gravitational PE. You are correct that the energy stored as spring PE is given by .5kx^2 where x is measured from the unstretched position of the spring. But what they are really asking for is the work you must do on the spring to stretch it an additional distance, not the total work done on the spring. (Gravity does part of the work.)

Let's work it out in detail and maybe it'll be clearer:

The amount of stretch in the spring when at equilibrium = mg/k
The spring PE at equilibrium = .5k(mg/k)^2
The amount of stretch in the spring when pulled an addition distance x = mg/k + x
The spring PE at that point = .5k(mg/k + x)^2

Measured from equilibrium, the change in spring PE = .5k(mg/k + x)^2 - .5k(mg/k)^2
Measured from equilibrium, the change in gravitational PE = -mgx

Measured from equilibrium, the total change in potential energy
= .5k(mg/k + x)^2 - .5k(mg/k)^2 -mgx = .5kx^2

Make sense?