T∆S implies that heat flows which way?
The fact that the book is using $\Delta G$ as the relevant state variable for free energy implies this is being carried out at constant T and p.  So, proceeding:


  "However, this implies that we are getting usable work from increases in entropy...."


Yes, that's correct.  The positive $\Delta S$ lowers $\Delta G$


  "However, this implies that we are getting usable work from increases in entropy – meaning that the muscles are somehow absorbing heat energy from the environment."


No, that's incorrect.  The heat flow here is purely due to $\Delta H$, and it is flowing out of the system.   

What you're probably thinking is that, at constant T (which we have here), $$\Delta S = \frac{q}{T},$$
and thus, since $\Delta S &gt;0 $,  q must be $&gt; 0$ as well, i.e., that the  positive $\Delta S$ means heat is flowing into the system.

This is wrong.  It is not the case that $\Delta S = \frac{q}{T}$.  Rather:

$$\Delta S = \frac{q_{rev}}{T},$$

and this reaction is not happening reversibly.  Instead,  $\Delta S$  is determined from the difference between $S_{products}$ and  $S_{reactants}$, and each of these individual values are calculated from:

$$S(T') =\int_{0}^{T'} dS= \int_{0}^{T'}\frac{\text{đ}q_{rev}}{T}$$

[For more on this, see my answer here:  Infinite Increase in Entropy when Energy added to Absolute Zero ]

In summary, in a chemical reaction at constant pressure (and with no non-pV work), 

$$\text{đ}q_{sys}=dH_{sys}$$

Thus it is the sign of $\Delta H$, not $\Delta S$, that determines the direction of heat flow.

Finally, the answer to (d) is that, as thermal energy is generated by the reaction, it heats up the system, and the 2nd law of thermodynamics says that thermal energy naturally flows from hotter to cooler regions.  So there is heat flow into the surroundings until the reaction mixture returns to the temperature of the surroundings.