= Solution
Assume the known variances are positive and write $\bar x=n^{-1}\sum_ix_i$. The <likelihood> times the normal <prior distribution> is proportional to
$$
\exp\left\{-\frac12\left[\frac1{\sigma^2}\sum_i(x_i-\theta)^2+
\frac{(\theta-\mu)^2}{\tau^2}\right]\right\}.
$$
Collecting the quadratic and linear terms in $\theta$ completes the square with precision $v^{-1}=n/\sigma^2+1/\tau^2$ and mean $m=v(n\bar x/\sigma^2+\mu/\tau^2)$. Thus <normal-normal conjugacy> gives
$$
\boxed{\theta\mid x_1,\ldots,x_n\sim N(m,v),\qquad
v=\frac{\sigma^2\tau^2}{\sigma^2+n\tau^2},\quad
m=\frac{n\tau^2\bar x+\sigma^2\mu}{\sigma^2+n\tau^2}.}
$$
For <quadratic loss>, the posterior risk of reporting $a$ is $\mathbb E[(a-\theta)^2\mid x]=v+(a-m)^2$, uniquely minimized at $a=m$. For <absolute-error loss>, differentiating the posterior risk gives $2F_{\theta\mid x}(a)-1$, so its minimizer is a posterior median. The normal posterior is symmetric about $m$ and has a strictly increasing distribution function, hence that median is uniquely $m$. Therefore \b[both optimal point estimates equal $m$], the <posterior mean>.
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