Consequences of high power for our belief in the null hypothesis? Suppose that we choose the size of the test to be $\alpha = 0.05$, and based on our sample size $n$ and magnitude of the absolute value of the estimate, we determine that the test's power (i.e., 1 - Prob(Type 2 Error)) is $99.9\%$. 
If I then fail to reject $H_0=0$ against $H_a \not= 0$, what consequences does this have for our belief that $H_0$ is correct, assuming that we have the correct null hypothesis distribution of the test statistic? 
What if the power was $50\%$?
 A: When you calculate power it is for a specific alternative value, so failing to reject the null could mean that the null is true, or it could be that it is false, just not as strong as the alternative value.
What is much more meaningful is to look at the confidence interval to see what the plausible/reasonable values the true parameter could be.
Often it is best to think about not just about what the null value would be but what the region of values would be that are practically equivalent to the null (not equal to the null, but close enough that we would not care) vs. the region of practical importance.  Then to see where the confidence interval lies.  Even if the interval does not include the null, but is completely in the region of "who cares" then that tells us something.  If the interval only includes values of interest then that tells us something else.  The big problem comes when the interval contains both values of practical importance and the null value, then our results are indeterminant, the result could be nothing or it could be important.  Better than worrying about power for a specific alternative is to design the study so that the confidence interval is too narrow to include both the null value and the smallest important difference (this will result in high power, but is a better way to think of things).
A: I will answer this question from the Bayesian perspective. Consider Bayes' theorem:
$$
P(H_0\mid S) = \frac{P(S \mid H_0)}{P(S)}P(H_0)
$$
where $S$ is a significant result. We can write an analogous expression for $H_A$. Then, dividing the two expressions, we get
$$
\frac{P(H_A\mid S)}{P(H_0\mid S)} = \frac{P(S\mid H_A)}{P(S\mid H_0)}\times\frac{P(H_A)}{P(H_0)}
$$
or 
$$
\mbox{Posterior odds} = \frac{\mbox{Power}}{\mbox{Type I error rate}}\times\mbox{Prior odds}.
$$
The term in the middle is called the Bayes factor, and represents the evidence in the data, or the effect on our belief. In particular, the effect of a significant result on our belief (in terms of relative odds of the alternative to the null) is exactly the power over the type I error rate, IF we only know that the result is significant (other facts, for instance the exact $p$ value, might change this assessment). This a significant result with $\alpha=0.05$ and power of 0.8 has an effect of .8/.05=16 on our relative beliefs.
Now for a nonsignificant effect, $N$,
$$
\frac{P(H_0\mid N)}{P(H_A\mid N)} = \frac{P(N\mid H_0)}{P(N\mid H_A)}\times\frac{P(H_0)}{P(H_A)}
$$
which is
$$
\mbox{Posterior odds} = \frac{1-\alpha}{\beta}\times\mbox{Prior odds}.
$$
The Bayes factor is thus $(1-\alpha)/\beta$. Again, this is the effect that the nonsignificant result has on our (relative) beliefs, IF all we know is that the result is nonsignificant.
Taking your example numbers, if $\alpha=.05$ and $1-\beta=.999$, then
$$
\frac{1-\alpha}{\beta} = \frac{.95}{.001} = 950,
$$
that is, the relative evidence in favor of the null hypothesis is enough to shift your odds by a factor of 950. This is extremely strong evidence.
If the power is .5, then
$$
\frac{1-\alpha}{\beta} = \frac{.95}{.5} = 1.9,
$$
which is extremely weak evidence (keep in mind 1.0 is no evidence either way).
