= Solution
The <spectral theorem for real symmetric matrices> identifies the sum of the $k$th powers of the <eigenvalues> with the <trace> of the $k$th <matrix> power. Therefore the <moment> of the <empirical spectral measure> is
$$
\boxed{m_k(n)=n^{-1-k/2}\operatorname{tr}(M^k)
=n^{-1-k/2}\sum_{i_1,\ldots,i_k=1}^n
M_{i_1i_2}M_{i_2i_3}\cdots M_{i_ki_1}.}
$$
For odd $k$, every summand has zero <expectation> by part (a)(ii), so $\boxed{\mathbb E[m_k(n)]=0}$ for every $n$.
For the second <moment>, use $M_{ji}=M_{ij}$. There are $n$ diagonal entries with <variance> two and $n(n-1)$ ordered off-diagonal entries with <variance> one. Their <means> vanish, giving
$$
\mathbb E[m_2(n)]
=n^{-2}\sum_{i,j}\mathbb E[M_{ij}^2]
=\frac{2n+n(n-1)}{n^2}
=1+\frac1n.
$$
Hence $\boxed{\lim_{n\to\infty}\mathbb E[m_2(n)]=1}$.
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