= Solution
Differentiate the <Helmholtz free energy> at fixed couplings. With $\lambda=\lambda_+$ from <1/b/solution>, the <entropy> per <spin> is
$$
s=-\frac{\partial f}{\partial T}=\frac{k_B}{2}\left(\log\lambda-\beta\frac{\partial\log\lambda}{\partial\beta}\right).
$$
It is particularly useful to write
$$
\lambda=e^{\beta(2\bar J-J)}+e^{-\beta(2\bar J+J)}+2e^{\beta J}.
$$
At $\bar J=J$, this is $3e^{\beta J}+e^{-3\beta J}$, so $f=-J/2-(\log3)/(2\beta)+o(\beta^{-1})$. For fixed $\bar J>J$, the first exponential dominates alone, giving $f=(J-2\bar J)/2+o(\beta^{-1})$. Hence the <residual entropy> is
$$
\boxed{\lim_{T\to0}s=\begin{cases}(k_B/2)\log3,&\bar J=J,\\0,&\bar J>J.\end{cases}}
$$
The finite positive difference $\bar J-J$ must be kept fixed in the second limit; taking it to zero first gives the other answer.
At high <temperature>, $\lambda\to4$ and $\beta\partial_\beta\log\lambda\to0$, whence
$$
\boxed{\lim_{T\to\infty}s=k_B\log2.}
$$
All $2^{2L}$ configurations become equally probable because their <Boltzmann factors> tend to one. The two available states per <spin>, rather than the coupling values, determine this <entropy>.
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