Why is the inapproximability $2^{n^\epsilon}$?
The two theorems are equivalent. You can write
$$
3^{n^\delta} = 2^{\log_2 3 \cdot n^\delta} = 2^{n^{\delta + \frac{\log \log_2 3}{\log n}}} = 2^{n^{\delta + o(1)}}.
$$
In particular, for any $\delta' &gt; \delta$ there exists $N$ such that $3^{n^\delta} \leq 2^{n^{\delta'}}$ for all $n \geq N$. Conversely, $3^{n^\delta} \geq 2^{n^\delta}$.

Hardness of approximation results are only about "large enough $n$". Every problem can be solved exactly in polynomial time (even constant time) for all inputs of size at most $N$, for any fixed $N$.