= Solution
Write $R(\theta,\delta)=\mathbb E_\theta L(\theta,\delta(X))$ and $r(\pi,\delta)=\int R(\theta,\delta)\,d\pi(\theta)$. An <equaliser rule> has constant risk $R(\theta,\delta)=c$. An <Extended Bayes rule> has, for every $\varepsilon>0$, a proper prior $\pi$ such that $r(\pi,\delta)\le r^*(\pi)+\varepsilon$, where $r^*(\pi)=\inf_d r(\pi,d)$. If a rule has both properties, every competitor $d$ satisfies
$$
\sup_\theta R(\theta,d)\ge r(\pi,d)\ge r^*(\pi)\ge c-\varepsilon.
$$
Letting $\varepsilon\to0$ proves that its maximum risk $c$ is the <minimax risk>.
For the given <normal distribution> model, completing the square in prior times likelihood gives the posterior $N(m,H^{-1})$, with $H=h_0+nh$ and $m=(h_0m_0+h\sum_i x_i)/H$. Another completion of the square gives
$$
\mathbb E\!\left[e^{-k(a-\theta)^2/2}\mid X=x\right]=\sqrt{\frac H{H+k}}\exp\!\left[-\frac{kH}{2(H+k)}(a-m)^2\right].
$$
This is maximized uniquely at $a=m$. Consequently the <Bayes act> and posterior <posterior expected loss> are
$$
\boxed{a_B=m,\qquad \rho_B(x)=1-\sqrt{\frac{h_0+nh}{h_0+nh+k}}.}
$$
The posterior loss is independent of $x$, so averaging over the predictive distribution gives the same value for the <Bayes risk> of this <Bayes decision rule>.
The <sample mean> obeys $\bar X-\theta\sim N(0,(nh)^{-1})$, so its risk is the constant $c=1-\sqrt{nh/(nh+k)}$. Take any fixed $m_0$ and let the proper prior precisions $h_0\downarrow0$. The optimal <Bayes risks> just calculated tend to $c$, while the integrated risk of $\bar X$ is exactly $c$ for every prior. Thus its excess <Bayes risk> tends to zero: it is extended Bayes as well as equaliser. The preceding argument proves
$$
\boxed{\bar X\text{ is minimax, with risk }1-\sqrt{\frac{nh}{nh+k}}.}
$$
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