Let $A$ be the event that our randomly selected person is married. Let $B$ be the event that our randomly selected person is male.
We are told $Pr(A\mid B)=0.6$ and that $Pr(A\mid B^c)=0.5$
Further, we are told that $Pr(B\mid A)=2Pr(B^c\mid A)$
We know from earlier results that $Pr(B\mid A)+Pr(B^c\mid A)=1$, so this tells us what $Pr(B\mid A)$ is.
Now, let $Pr(B)=x$. We wish to find what $x$ is.
We know that $Pr(A\mid B)=\dfrac{Pr(B\mid A)Pr(A)}{Pr(B)}$ and that $Pr(A)=Pr(B)Pr(A\mid B)+Pr(B^c)Pr(A\mid B^c)$.
Plugging in values and completing the necessary algebra lets us find the value of $x$ from which we can now find the number of females in the company.
Let's do it the exact way first, then use your normal distribution approach.
Under the assumption that a person answering the call is equally likely to be male as they are likely to be female, then the random number $X$ of females in $n = 70$ answered calls is a binomial random variable, namely $$X \sim \operatorname{Binomial}(n = 70, p = 0.5), \\ \Pr[X = x] = \binom{70}{x} (0.5)^x (1 - 0.5)^{70 - x} = \binom{70}{x} (0.5)^{70}.$$ Therefore, $$\Pr[33 \le X \le 36] = \sum_{x=33}^{36} \binom{70}{x} (0.5)^{70} = (0.5)^{70} \left(\binom{70}{33} + \binom{70}{34} + \binom{70}{35} + \binom{70}{36} \right).$$ With the aid of a computer (although it can be tediously computed by hand), this is exactly $$\Pr[33 \le X \le 36] = \frac{26909546368186020357}{73786976294838206464} = 0.36469235791233373812\ldots.$$ This represents the precise probability, with no approximations.
Now let's use the normal distribution approach. The idea is to model the random variable $X$ with a suitable normal distribution whose mean and variance match the binomial mean and variance; i.e., $$X \sim \operatorname{Normal}(\mu = np, \sigma^2 = np(1-p)),$$ which gives $\mu = 35$, and $\sigma^2 = 17.5$ for $n = 70$ and $p = 0.5$. Then we standardize $X$: $$\Pr[33 \le X \le 36] \approx \Pr\left[\frac{33 - 35}{\sqrt{17.5}} \le \frac{X - \mu}{\sigma} \le \frac{36 - 35}{\sqrt{17.5}} \right] \approx \Pr[-0.478091 \le Z \le 0.239046],$$ where $$Z = \frac{X - \mu}{\sigma} \sim \operatorname{Normal}(0, 1)$$ is a standard normal random variable whose probabilities we can look up in a table. If we do, we find $$ \Pr[-0.478091 \le Z \le 0.239046] = \Phi(0.239046) - \Phi(-0.478091) \approx 0.594465 - 0.316293 \approx 0.278172.$$
Where did we go wrong? Why is this approximation so poor? The problem here is that we did not employ continuity correction. The approximation we used fails to capture the full probability mass at $X = 33$ and $X = 36$. To compensate, we must instead write $$\Pr[33 \le X \le 36] \approx \Pr[33 - 0.5 \le X \le 36 + 0.5],$$ because both endpoints of the interval are included, so we must enlarge the interval by $0.5$ in each direction. Then the revised bounds on the standardized normal is $$\Pr[-0.597614 \le Z \le 0.358569] \approx 0.364992,$$ and as you can see, this result is much closer to the exact value we showed above.
Where did your computation go wrong? Well, for one thing, your reasoning for the standard deviation $\sigma$ is incorrect, as I have explained. Next, you didn't use continuity correction. As we saw, this is needed when using a continuous probability distribution to approximate a discrete distribution, and the error that occurs without continuity correction is large in this case because the standardized endpoints $[-0.47, 0.23]$ are not far enough away from $0$, where much of the probability mass lies in a standard normal distribution.
Best Answer
You are told that the accounting department provides jobs for 12% of the males. In other words, among the males, $12\%$ are in the accounting department. In other other words, given that one is male, there is a $12\%$ chance he is employed by accounting: $$P(A|M) = .12$$
To get a better understanding, trying drawing a picture. Or, imagine a company outing in which the employees wear their department tags. If the employees are separated into male and female, then $12\%$ of the males will be wearing accounting tags.
You are not told the chance that one is male given that they are in accounting, $P(M|A)$. You need to calculate this.
Similar reasoning follows for females.