Here is my take for $a)$
From the question, I can find:
$$n=124, p = \frac{2}3, np = 82.667 \ge 10, nq = 41.133 \ge 10$$ $$u = np = 82.667, \sigma^2 = npq = 27.5556$$
Test of hypothesis (two-tail): $$ H_0 : p = \frac{2}3 \space \space vs \space \space H_a: p \not= \frac{2}3$$
and I used,
$$z = \frac{p - p_0}{\sqrt{p_0(1-p_0)/n}}$$
where $p_0 = \frac{2}3, p = \frac{80}{124}$ then i get $$z = -.508$$
Hence, P-value $$= 2P(z\ge -.508)$$ $$=1.478$$
but apparently, the correct answer is $0.6114$
What did I do wrong? Any help would be awesome!!
Quite simply, you've taken the wrong tail of the standard normal distribution. You should have calculated $$2 \Pr[Z \le -0.508] = 0.6114.$$ The reason is that if the test statistic you calculate is negative for your test, then you take the probability that $Z$ is less than or equal to this value; if the test statistic is positive, you take the probability that $Z$ is greater than or equal to this value.