• Laplace location family: recall here we have shown if ${\displaystyle X_1,\cdots ,X_n\stackrel{\mathrm{i}.\mathrm{i}.\mathrm{d}}{\sim }\mathrm{Laplace}(\theta )}$, ${\displaystyle S(X)=(X_{(1)},\cdots ,X_{(n)})}$ is minimal sufficient. However, it's not complete. Denote ${\displaystyle \mathrm{Median}(S),\bar{X}(S)}$ as the sample median and sample mean of ${\displaystyle S}$. Both of them are unbiased estimators for ${\displaystyle \theta }$, so ${\displaystyle f(S)=\mathrm{Median}(S)-\bar{X}(S)}$ has expectation ${\displaystyle 0}$. But ${\displaystyle f(S)\stackrel{\mathrm{a}.\mathrm{s}.}{\ne }0}$.
• Uniform scale family: ${\displaystyle X_1,\cdots ,X_n\stackrel{\mathrm{i}.\mathrm{i}.\mathrm{d}}{\sim }\mathrm{Uniform}[0,\theta ],\theta \in (0,+\mathrm{\infty })}$. Let ${\displaystyle T(X)=X_{(n)}}$. Then $${\mathbb{P}}_{\theta }(T\le t)=min{\{\frac{t}{\theta },1\}}^n=min\{{\left(\frac{t}{\theta }\right)}^n,1\},t>0,$$ so $$p_{\theta }(t)=n\frac{t^{n-1}}{{\theta }^n}\mathbf{1}\{t\le \theta \}.$$ Suppose ${\displaystyle f}$ s.t. $${\mathbb{E}}_{\theta }[f(T(X))]=\frac{n}{{\theta }^n}{\int }_0^{\theta }f(t)t^{n-1}\mathrm{d}t=0,\mathrm{\forall }\theta ,$$ we can differentiate to show that ${\displaystyle f(t)t^{n-1}=0}$, a.s. So ${\displaystyle T(X)=X_{(n)}}$ is complete.

Assume ${\displaystyle T(X)=X}$ (this won't affect completeness), and ${\displaystyle h(x)\equiv 1}$, so ${\displaystyle p_{\eta }(x)={\mathrm{e}}^{{\eta }^{\mathrm{T}}x-A(\eta )}}$, i.e. carnonical form. WLOG assume ${\displaystyle 0\in {\mathrm{\Xi }}^{\circ }}$ (otherwise we reparameterize).
Any measurable function ${\displaystyle f}$ can be decomposed as ${\displaystyle f(x)=f^{+}(x)-f^{-}(x)}$, where $$f^{+}(x)=max\{f(x),0\}\ge 0,f^{-}(x)=max\{-f(x),0\}\ge 0.$$
Assume $${\mathbb{E}}_{\eta }[f(X)]=\int (f^{+}-f^{-})p_{\eta }\mathrm{d}\mu =0,\mathrm{\forall }\eta \in \mathrm{\Xi },$$ i.e. $$\begin{array}{}\text{(*)}& \int {\mathrm{e}}^{{\eta }^{\mathrm{T}}x}{f}^{+}(x)\mathrm{d}\mu (x)=\int {\mathrm{e}}^{{\eta }^{\mathrm{T}}x}{f}^{-}(x)\mathrm{d}\mu (x).\end{array}$$
By assumption, ${\displaystyle \mathrm{\Xi }}$ contains an open neighborhood that contains ${\displaystyle 0}$, on which both integrals are finite. Denote ${\displaystyle c=\int f^{+}(x)\mathrm{d}\mu (x)\ge 0}$.

Assume ${\displaystyle S}$ is minimal sufficient, so any sufficient ${\displaystyle T}$, we have ${\displaystyle S(X)=f(T(X))}$. Now let ${\displaystyle \bar{T}(S(X))=\mathbb{E}[T(X)|S(X)]}$ (there is no ${\displaystyle \theta }$ under ${\displaystyle \mathbb{E}}$ because of sufficiency), and let ${\displaystyle g(t)=t-\bar{T}(f(t))}$. Then $$\begin{array}{rl}& {\mathbb{E}}_{\theta }[g(T(X))]={\mathbb{E}}_{\theta }T(X)-{\mathbb{E}}_{\theta }[\bar{T}(S(X))]\\ =& {\mathbb{E}}_{\theta }T(X)-{\mathbb{E}}_{\theta }[\mathbb{E}[T|S]]=0,\end{array}$$ the last equality is by tower property. So ${\displaystyle g(T(X))\stackrel{\mathrm{a}.\mathrm{s}.}{=}0}$, so ${\displaystyle T\stackrel{\mathrm{a}.\mathrm{s}.}{=}\bar{T}}$. This means we can calculate ${\displaystyle T}$ from ${\displaystyle \bar{T}}$, which is from ${\displaystyle S}$, so ${\displaystyle T}$ is also minimal sufficient.

We want to prove for any ${\displaystyle A,B,\theta }$, $${\mathbb{P}}_{\theta }(V\in A,T\in B)={\mathbb{P}}_{\theta }(V\in A){\mathbb{P}}_{\theta }(T\in B).$$
Define ${\displaystyle q_A(T(X))={\mathbb{P}}_{\theta }(V\in A|T)}$, ${\displaystyle p_A=\mathbb{P}(V\in A)}$ (${\displaystyle V}$ is ancillary, so no ${\displaystyle \theta }$ here), so $${\mathbb{E}}_{\theta }[q_A(T)-p_A]=p_A-p_A=0,\mathrm{\forall }\theta .$$
By completeness of ${\displaystyle T}$, this implies ${\displaystyle q_A(T)\stackrel{\mathrm{a}.\mathrm{s}.}{=}{p}_A}$. Then $$\begin{array}{rl}& {\mathbb{P}}_{\theta }(V\in A,T\in B)=\int q_A(t)\mathbf{1}\{t\in B\}\mathrm{d}{P}_{\theta }^{\mathrm{T}}(t)\\ =& \int p_A\mathbf{1}\{t\in B\}\mathrm{d}{P}_{\theta }^{\mathrm{T}}(t)={\mathbb{P}}_{\theta }(V\in A){\mathbb{P}}_{\theta }(T\in B).\end{array}$$

• We first assume ${\displaystyle {\sigma }^2}$ is known. Now ${\displaystyle \mathcal{P}}$ is a one-parameter full-rank exponential family with complete sufficient statistic ${\displaystyle \bar{X}}$. Moreover, ${\displaystyle S^2}$ is ancillary, since $$S^2=\sum _{i=1}^n(Z_i-\bar{Z}{)}^2,Z_i=X_i-\mu ,$$ and ${\displaystyle Z_1,\cdots ,Z_n\stackrel{\mathrm{i}.\mathrm{i}.\mathrm{d}}{\sim }\mathcal{N}(0,{\sigma }^2)}$ is known.
  So applying Basu's Theorem, we have ${\displaystyle \bar{X}\perp \perp S^2}$.
• Since ${\displaystyle {\sigma }^2}$ is arbitrarily picked, we have the conclusion hold for all ${\displaystyle \mu ,{\sigma }^2}$.