A company produces millions of 1-pound packages of bacon every week. Company specifications allow fo

We can eliminate out of hand the obviously incorrect choices A, C, and D, leaving only B and E as possibly correct.
A is incorrect, because we know that just because the point estimate is above the hypothesized proportion, that does not imply that the variability or uncertainty in that estimate is small enough to state with a high degree of confidence that the true proportion exceeds the hypothesized proportion.
Another way to think about it is that if you flip a fair coin 10 times, there's quite a reasonable chance you may get at least 6 heads (more than 1 in 3 chance), despite the coin being perfectly fair. Yet, if you flip the same coin 1000 times and get 600 or more heads, this is extremely improbable (odds less than 1 in 7 billion), despite both sample proportions being 0.6=6/10=600/1000.
C is incorrect because the conclusion is opposite of what is required by the p-value; i.e., if $p>\mathrm{\alpha <!--\; \alpha \; -->}$, you fail to reject the null hypothesis, whereas the answer choice states that you would reject (Yes) when 0.064>0.05.
D is incorrect for the same reason as C: the conclusion is opposite of what is required. If $p>\mathrm{\alpha <!--\; \alpha \; -->}$, you would reject the null hypothesis, whereas the answer choice states that you would fail to reject (No) when 0.032<0.05.
This leaves only B and E as choices. Which to choose depends on which p-value is correct. Our hypothesis is$H_0:p=p_0=0.03\text{vs.}{H}_a:p>p_0=\mathrm{0.03.}$
The test statistic is
$Z\mid <!--\; \mid \; -->H_0=\frac{\stackrel{^<!--\; ^\; -->}{p}-<!--\; -\; -->p_0}{\sqrt{p_0(1-<!--\; -\; -->p_0)/n}}\sim <!--\; \backslash sim\; -->\mathrm{Normal}<!--\; \; -->(0,1).$
This much you have stated correctly, but the calculation should be
$Z\mid <!--\; \mid \; -->H_0=\frac{0.04-<!--\; -\; -->0.03}{\sqrt{0.03(0.97)/1000}}\approx <!--\; \backslash approx\; -->\frac{0.01}{0.00539444}\approx <!--\; \backslash approx\; -->\mathrm{1.85376.}$
The p-value for this one-sided test is the probability
$Pr[Z>1.85376]\approx <!--\; \backslash approx\; -->0.0318868,$
which matches the 0.032 p-value in answer choice B.
For your second question, you have insufficient information to compute a $p$-value. What you must do is make a decision based on the information provided. Since the two-sided 90% confidence interval contains the null mean 100, you know that the two-sided hypothesis test at α=1-0.9=0.10 must not be able to reject $H_0$ in favor of $H_a$: the uncertainty in your point estimate is too large to rule out with 90% confidence that the true mean is different than 100, because the CI contains 100. This also means that p-value cannot be less than $\mathrm{\alpha <!--\; \alpha \; -->}=0.10$, otherwise you would reject $H_0$.
So the correct answer is D.