The confidence interval with $\frac{1}{\sqrt{n}}$ is based on the same idea as the confidence interval with $1.96\sqrt{\frac{p(1-p)}{n}}$ but is more "conservative", in the sense that it is larger. The reason for that is that the function

$$f(x) = x \left(1-x\right), x \in [0,1]$$

can be shown (with elementary calculus) to be maximal for $x= \frac{1}{2}$. Thus

$$1.96\sqrt{\frac{p(1-p)}{n}} \leq 1.96 \sqrt{ \frac{1}{4n}} \approx \frac{1}{\sqrt{n}}$$

The confidence interval with $1.96\sqrt{\frac{p(1-p)}{n}}$ will be close to the one with $\frac{1}{\sqrt{n}}$ for $p \approx 1/2$ but since in none of your problems is that the case, I would probably use the classical $1.96\sqrt{\frac{p(1-p)}{n}}$ CI.

For confidence intervals we always use the form $\hat{p} \pm z\sqrt{\frac{p(1-p)}{n}}$. For the 95% confidence interval $z=1.96$

Since the term $z\sqrt{\frac{p(1-p)}{n}}$ depends on $p$ there are some proportions which have bigger confidence intervals. The worst case scenario is where $p=0.5$, this has the most variation which is why the confidence interval is larger.

Before we collect any data we might want to estimate how good the results might be. Since we don't have data for the proportion we can just use the worst case scenario. For the worst case scenario the accuracy indicated by the confidence interval is $z\sqrt{\frac{0.5(1-0.5)}{n}}$

For the 95% confidence interval where $z=1.96$ this simplifies to $\frac{0.98}{\sqrt{n}}$

This is where your idea of $\frac{1}{\sqrt{n}}$ comes from. The constant in the numerator is $0.98$ which is close to $1$ but this is only for the 95% confidence interval, other levels of confidence will have other values.

In your questions you have data for every case so you won't need to use the worst case scenario. However, it's still useful to remember for times when you are planning an experiment and you want to see if it is feasible before collecting any data.