Bags of concrete mix labeled as containing 100 lb have a population mean weight of 100 lb Bags of concrete mix labeled as containing 100 lb have a population mean weight of 100 lb and a population standard deviation of 0.5 lb
a) Find probability that the mean weight of a random sample of 50 bags is less than 99.9 lb
This part was easy for me, i used the normal distribution model and got 0.0228 as the probability
b) Find the 5th percentile of the sample means
I don't know about this part, i know i'm supposed to use the normal distribution model but im not sure how to get the correct answer here
c) find the sample size so that no less than 99% of the sample means are within 0.05 of the true population mean
 A: Part (a). If $X_1, X_2, \dots, X_{50}$ is a random sample from $\mathsf{Norm}(\mu = 100,\, \sigma = 0.5),$ then $\bar X \sim \mathsf{Norm}(\mu = 100,\, \sigma = 0.5/\sqrt{50}).$
If you want to compute $P(\bar X \le 99.9),$ you can use the following approach, which allows you to use printed normal CDF tables for the last step:
$$P(\bar X \le 99.9) = 
P\left(\frac{\bar X - \mu}{\sigma/\sqrt{n}}\le \frac{99.9 - 100}{0.5/\sqrt{50}}\right) = P(Z \le -1.414) =\, ??$$
You will have to round -1.414 (or interpolate) to use most printed tables.
The exact answer to four places from R statistical software is 0.0786, as
Commented by @GTonyJacobs. But your value from tables may be something like 0.0793, depending on the table you use (and exactly how).
pnorm(99.9, 100, .5/sqrt(50))
## 0.0786496

Part (b). The 5th percentile of the distribution of $\bar X$ is the number $q$ such that
$P(\bar X \le q) = 0.05.$
To find this from normal tables, write
$$P(\bar X \le q) = 
P\left(\frac{\bar X - \mu}{\sigma/\sqrt{n}}\le \frac{q - 100}{0.5/\sqrt{50}}\right) = 0.05.$$
In your normal table, find closest available number in the body of the table 
to 0.05. Then find the value $c$ such that $P(Z \le c) = 0.05.$ Because
$c$ will be negative (depending on the style of printed table you are using),
you may need to find the value that cuts 0.05 from the upper tail of the
standard normal distribution, and then use symmetry to find $c.$ When you get $c$, Set $c = \frac{q-100}{0.5/\sqrt{50}},$ and solve for $q.$ The exact
value from software is shown below, but the potential for small rounding errors using tables is more serious here than in part (a), so do not expect to get exactly the value shown using tables.
qnorm(.05, 100, .5/sqrt(50))
## 99.88369         # exact q
qnorm(.05)
## -1.644854        # exact c

Part (c). Here I will give only a hint, because I don't know what
you have been studying recently.
A 99% confidence interval (CI) for $\mu$ is of the form
$$\bar X \pm 2.576\frac{\sigma}{\sqrt{n}}.$$
This is derived from the probability 
$$P\left(\frac{|\bar X - \mu|}{\sigma/\sqrt{n}} \le 2.576\right) = .99.$$
The 'margin of error' of the CI is $E = 2.576\sigma/\sqrt{n}.$
Plug in known quantities and solve for $n;$ if $n$ is not an integer, round up.
I got something like $n=664.$
If you have a t-table in your textbook, you may want to know that the
bottom row of that table (perhaps marked: 'Inf, $\infty,$ or 'Norm')
is for the standard normal distribution. You may be able to get the z-value 2.576 from there. About the best you can do from printed normal tables (without interpolation) is 2.58. In answering, you should use only
values you can show how to get.
In the figure below, red dotted lines are relevant to part (a) and dashed cyan lines to part (b). In the normal probability plot, purple lines are at
$\pm 2.576.$

Note: I don't think the Chebyshev bound suggested in another answer to your question
is sufficiently accurate to get the answer you are supposed to provide.
A: Hint:
b)
$P(Z<=\frac{ \bar X - 100}{\frac{.05}{\sqrt{50}}}) = 0.05$
$\frac{ \bar X - 100}{\frac{.05}{\sqrt{50}}} = \Phi^{-1}(0.05)$
c) $P(|X-100|\le .05 ) \gt 0.99$
Using Chebychev's inequality
$P(|X-100|\ge .05 ) \le .01$
$P(|X-\mu| \ge \frac{k\sigma}{\sqrt{n}}) \le \frac{1}{k^2}$
Now equate these two expressions
$\frac{k\sigma}{\sqrt{n}} =.05$
But $ \frac{1}{k^2} = .01$
$k = 10$
$\sqrt{n} = \frac{10\times 0.5}{.05}$
$\sqrt{n} = 100$
$n = 10000$
