This probably isn't going to be in time for your lab tomorrow, but I'll attempt to shed some light on this:
In this sort of fluid-structure interaction with an oscillating/vibrating body, there are generally three fluid effects that need to be considered: added mass, added stiffness and added damping.
Added mass is basically the fluid's mass adding to that of the oscillator, because fluid has to be accelerated to move it out of the way of the moving object.
Added stiffness is fluid effects adding to the stiffness of the oscillator (I'll expand on that shortly).
Added damping is extra damping of the oscillator, caused by viscous friction of the object as it moves through the fluid.
The first two will tend to change the resonant frequency of a mechanical oscillator, as this is usually given by $\sqrt{\frac{k}{m}}$. Added damping will obviously increase the damping of the system and cause oscillations to subside more quickly, as it helps to dissipate energy.
So, in terms of the effect of the fluid on the spring constant 'k' in your experiment, it's the added stiffness that you should be looking at. So, what behaviour of a fluid will provide an opposing force that's proportional to the amount of displacement? The answer is buoyancy with a free surface. If an object is partially submerged, then the buoyancy force will increase as it submerges further and reduce as it lifts back up out of the fluid. If the object is completely submerged, then the buoyancy won't change with vertical displacement, so then there shouldn't be any added stiffness.
So, my prediction is that k will be higher if the object is partially submerged and will be unchanged if the object is completely submerged.
Conceptually, the reason they are attached to walls is to remove rigid body motion from consideration. When they are tethered to the wall, there is no translation or rotation about the center of mass of the system.
So, when you break the connections to the walls and they are now free to drift through space, the number of admissible solutions increases.
If you connect one mass to the wall and leave the other end unconnected, then you remove the rigid body modes because it cannot rotate or translate and you're back to a similar, but different, problem as when both are connected to walls.
The equations of motion are easily derived for 3 bodies in both conditions when one draws the free body diagram (if deriving from a Newtonian approach) or through the application of Lagrange's equation or Hamilton's principle, depending on your choice of weapon.
But it's not that it's harder or impossible to have the three masses floating in space, it just adds more admissible modes to the solution which may or may not obscure the intent of the example/problem.
Best Answer
No, the spring constant of a simple spring doesn't change. Here is what is happening: With zero added mass, the spring is trying to collapse into a shorter length but cannot because the coils are interfering with each other. They are exerting forces, keeping the spring longer than it would be if zero force was acting on each coil. The addition of the 80 g mass in earth's gravitational field is apparently enough to separate the coils. So calculating the spring constant with the first length change gives in inaccurate value.
Adding more mass will (in gravity) will give an accurate value of the force change versus the length change: $$\Delta \vec{F} = -k\Delta \vec{x}$$.