The precise version of the question was answered in the affirmative in the paper "Extremes, Extreme Spacings, and Tail Lengths: An Investigation for Some Important Distributions," by Mudholkar, Chaubey, and Tian (Calcutta Statistical Association Bulletin 61, 2009, pp. 243-265). (Unfortunately, I haven't been able to find an online copy.)
Let $X_{i:n}$ denote the $i$th order statistic from a random sample of size $n$. Let $S_{n:n} = X_{n:n} - X_{n-1:n}$, the rightmost extreme spacing. The OP asks for $E[S_{n:n}]$ when sampling from a normal distribution.
The authors prove that, for an $N(0,1)$ distribution, $\sqrt{2 \log n}$ $S_{n:n}$ converges in distribution to $\log Z - \log Y$, where $f_{Z,Y}(z,y) = e^{-z}$ if $0 \leq y \leq z$ and $0$ otherwise.
Thus $S_{n:n} = O_p(1/\sqrt{\log n})$ and therefore converges in probability to $0$ as $n \to \infty$. So $\lim_{n \to \infty} E[S_{n:n}] = 0$ as well. Moreover, since $E[\log Z - \log Y] = 1$, $E[S_{n:n}] \sim \frac{1}{\sqrt{2 \log n}}$. (For another argument in favor of this last statement, see my previous answer to this question.)
In other words, (2) is more surprising.