The short answer is, it doesn't really mean much of anything, because it's not meaningful to compare the Gibbs free energy of two systems unless they're at the same temperature. To show you why, I'll quickly run you through the derivation of the Gibbs free energy and point out where the constant temperature assumption comes in.
The second law of thermodynamics tells us that the total entropy of a system, plus the entropy of its surroundings, must be non-decreasing. I'll write this as $\Delta S_\text{total} = \Delta S + \Delta S_\text{surroundings} \ge 0$.
Suppose we have a system (of any kind) in contact with an environment that remains at constant temperature and pressure. This system undergoes a process (of any kind) that changes its internal energy by $\Delta U$ (with positive sign meaning the system's energy increases) and its volume by $\Delta V$ (positive sign meaning it does work on the environment). We know that the system has done an amount of work $p\Delta V$ on the environment, so we can say that $\Delta U = Q - p\Delta V$, with $Q$ an amount of heat transferred from the surroundings to the system (or in the opposite direction if it's negative).
Since the surroundings are at a constant temperature $T$, they must have lost an amount of entropy equal to $Q/T = \frac{\Delta U}{T} + \frac{p\Delta V}{T}$. So we have
$$\Delta S_\text{total} = \Delta S - \frac{\Delta U}{T} - \frac{p\Delta V}{T}.\qquad\qquad(i)$$
This quantity, as we know from the second law, must always seek a maximum, as long as the temperature and pressure remain constant.
Now, up to this point the assumption of constant temperature hasn't really been necessary. If we assume that $\Delta U$ etc. are small we can replace them with differentials to get $dS_\text{total} = dS - pdV/T - dU/T$, which we can integrate if we know how $p$ and $T$ depend on $U$ and $V$.
However, for reasons that always struck me as somewhat arbitrary, the tradition in physics is to multiply equation $(i)$ by $-T$ to get
$$\Delta G \stackrel{\textit{def}}{=} -T\Delta S_\text{total} = \Delta U + p\Delta V - T\Delta S,$$
which must now be minimised because of the change of sign. I suppose this is done in order to put the quantity into energy units, which physicists and chemists tend to find more intuitive than entropy units. But it comes at a price - the transformation from maximising $\Delta S_\text{total}$ to minimising $-T\Delta S_\text{total}$ only works if $T$ is a constant, and if you violate this assumption then $\Delta G$ can either be positive or negative for a spontaneous process, so its sign no longer really means anything.
To work out $\Delta S_\text{total}$ for your system you would have to integrate $dS_\text{total}$ as described above, and to do that you would have to know the heat capacity and thermal expansion coefficient of steam (as functions of $T$ and $p$, though I'd guess they're pretty constant over that range) in addition to the data you posted. This would certainly come out positive, and a lower value would mean a more efficient turbine, with $\Delta S_\text{total}$ approaching zero in the reversible limit.
There is no contradiction here. The change in Gibbs free energy tells you the maximum extractable work and at constant teperature and pressure $dG = 0$. The conclusion here is that you can extract no work while maintaining a fixed temperature and pressure.
Why is this? Well you have 5 variables in your thermodynamic system $U, p, V, T$ and $S$. We can replace $U$ with $G$ in our list using $G = U +pV - TS$. Now we know from the fundamental relation that
\begin{align}
G &= G(p,T)\\
V &= \left(\frac{\partial G}{\partial p}\right)_T\\
S &= \left(\frac{\partial G}{\partial T}\right)_p
\end{align}
So we have 5 variables and 3 constraint, leaving us with 2 degrees of freedom in our system. If we now impose another 2 constants, say by fixing $p$ and $T$, we have fully defined the state of our system. This means under these conditions we can extract no work because there is no other state we can move to in the transition.
If we do want to extract some useful work from our system there are 2 things we can do; either we can relax one of our constraints, say by allowing temperature to vary, or we can introduce additional physical degrees of freedom to the system, for example by allowing particle number to vary, and modify the expression for the change in $G$ accordingly, as @higgsss describes in they're answer.
Best Answer
Consider a change of the energy of a system system, given the natural variable $S,V$ (entropy and volume): $$dE = \delta W + \delta Q$$ Where $W$ is the mechanical work (volume change) and $Q$ is the energy changes via heat flux.
We know that the mechanical work exchanged from the system with the environment is link with volume change and pressure i.e. $W(P,V)$ and the thermal exchanges are dependant on the entropy and temperature i.e. $Q(T,S)$.
Now impose a constraint on your system imposing constant temperature $T_0$. Obviously $\delta W=0$ during transformation your system will do in order to maximize it's Helmoltz free energy as the volums stays constant. Only heat exchanges will occur and $A$ will go to its maximum at equilibrium. What's left is the mechanical work you can extract from your system through volume changes (expansion work)!
Now impose a constraint on your system regarding the pressure and the temperature. Reaching the new equilibrium will force your system to go through temperature changes and volume changes until it reaches equilibrium so your system can't heat up anything (due to the reach of equilibirum temperature with heat bath) and you cannot extract mechanical work (due to the reach of equilibrium with pressure bath). What you can extract though is chemical work through electricity (non expansion work).
You actually could already do this at constant volume though so you see that in the first case, you can still extract chemical and mechanical work (maximum possible work in your book), defined by Helmoltz Free energy, but in the second case, as in the equilibrium your system as already given it's mechanical work in order to reach equilibrium, you can only extract chemical work (non expansion work) represented by Gibbs free energy. That's why the later is often used to describe chemical reactions at constant temperature and volume.
I hope this helps.