= Solution
An <estimator> $\hat\theta_n$ is <consistent> if, for every parameter value $\theta$ and every $\epsilon>0$, $\mathbb P_\theta(|\hat\theta_n-\theta|>\epsilon)\to0$.
For the proposed root criterion, fix $\epsilon>0$. <Convergence in probability> at the two points gives
$$
\mathbb P(S_n(\theta_0-\epsilon)<0<S_n(\theta_0+\epsilon))\to1.
$$
On this event the <intermediate value theorem> gives a zero between the two points. Since the zero is unique, it is $\hat\theta_n$. Therefore \b[$\boxed{\hat\theta_n\xrightarrow{P}\theta_0}$], without needing uniform convergence of $S_n$.
For independent $N(\theta_0,1)$ observations, the <log-likelihood> has derivative $n(\bar X_n-\theta)$ and negative second derivative $-n$. Its unique maximizer is $\hat\theta_n=\bar X_n$. Apply the criterion to $S_n(\theta)=\theta-\bar X_n$: the <strong law of large numbers> gives the limit $S(\theta)=\theta-\theta_0$, with the required signs. Hence the <maximum-likelihood estimator> is consistent.
In the bivariate model the log-likelihood, apart from constants, is
$$
\ell(\mu,\sigma^2)=-n\log\sigma^2-\frac1{2\sigma^2}\sum_{i=1}^n[(X_{1i}-\mu_i)^2+(X_{2i}-\mu_i)^2].
$$
Minimizing each squared pair gives $\boxed{\hat\mu_i=(X_{1i}+X_{2i})/2}$. Maximizing the resulting likelihood in the variance gives
$$
\boxed{\hat\sigma^2=\frac1n\sum_{i=1}^ns_i^2=\frac1{4n}\sum_{i=1}^n(X_{1i}-X_{2i})^2.}
$$
Each difference is $N(0,2\sigma^2)$, independently of the others and independently of its unknown mean parameter. Thus the <strong law of large numbers> gives $\hat\sigma^2\to\sigma^2/2$ in probability, not $\sigma^2$. This is the <Neyman-Scott incidental parameter problem>: the number of nuisance means grows with the sample size, but each mean is estimated from only two observations. The usual fixed-dimensional likelihood consistency theory therefore does not apply. The corrected estimator $2\hat\sigma^2$ is consistent.
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