= Solution
Write $Y_i=X_i^T\beta+\varepsilon_i$, where conditionally $\mathbb E(\varepsilon_i|X_i)=0$ and $\operatorname{Var}(\varepsilon_i|X_i)=\sigma^2v(X_i)$. Then
$$
\widehat\beta(\Sigma)-\beta=
\left(\frac1n\sum_i\frac{X_iX_i^T}{v(X_i)}\right)^{-1}
\frac1n\sum_i\frac{X_i\varepsilon_i}{v(X_i)}.
$$
The law of large numbers sends the <matrix> to
$$
A=\mathbb E\left[\frac{XX^T}{v(X)}\right],
$$
while the <vector> tends to zero, proving consistency. The central <limit> theorem and Slutsky's theorem give
$$
\sqrt n(\widehat\beta(\Sigma)-\beta)\Rightarrow N(0,\sigma^2A^{-1}).
$$
Similarly, with $B=\mathbb E(XX^T)$ and $C=\mathbb E(v(X)XX^T)$,
$$
\sqrt n(\widehat\beta(I)-\beta)\Rightarrow
N(0,\sigma^2B^{-1}CB^{-1}).
$$
These conclusions require the displayed <matrices> to be finite and nonsingular.
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