The answer is yes and no: You only really need to use it if the area under the normal curve will change substantially with a 0.5 unit change in the location on the x axis. In the normal approximation, your mean and variance are large, so it is not necessary.
In the second case, the underlying distribution is not defined over the integers, but over a discrete, finite set of fractions much smaller than the integers, hence again the proper correction in that case will not affect the answer.
First of all, you did the second calculation incorrectly. You should have had $F(0.0032)-F(-0.0032)$. If you do this, you will see that the result of the second calculation is approximately 50 times smaller than the result of the first. (This comes from continuity of the normal PDF.) This is to be expected because you're using an interval of length $0.02=1/50$ instead of an interval of length $1$. That is, a normal approximation of $50 (F(0.0032)-F(-0.0032))$ would give you a very similar result to the first calculation.
The rest of this is a somewhat subtle discussion about this issue, which may not be quite suitable for a beginning statistics student.
The meaningful thing is that the binomial CDF $F_b$ and the appropriate normal CDF $F_n$ are close to one another. The continuity correction is used to relate this to the binomial PMF $f_b$. The problem is that $f_b$ can be represented in any number of different ways in terms of $F_b$.
Specifically, for any two real numbers $x,y \in [0,1)$, $f_b(m)=F_b(m+x)-F_b(m-1+y)$. To approximate $f_b(m)$ using $F_n$, we choose $x$ and $y$ and then approximate these values of $F_b$ using the corresponding values of $F_n$.
But it is not good enough to just get an acceptable approximation of these values of $F_b$, because when $n$ (the number of binomial trials) is large, these two numbers are very close. As a result small relative errors in our approximations of $F_b(m+x)$ and $F_b(m-1+y)$ can cause large relative errors in the difference. (As an example, consider $x=1.001$, $y=1$, and attempting to approximate $x-y$ when replacing $x$ with $z=1.0015$. Here $z$ and $x$ are relatively quite close together but $x-y$ and $z-y$ are relatively quite far apart.)
It turns out that we can only make this error even moderately small if $x=y$. More precisely, if $x \neq y$ then our value will be off by a factor of about $x-y+1$, even as $n \to \infty$. This is essentially because the values of the binomial variable are separated by intervals of length $1$; the precise details involve analytical tricks that are beyond the scope of both calculus and elementary statistics.
Moreover the error is minimized in an appropriate sense when $x=y=1/2$. Nevertheless, if $X$ is $Bin(n,p)$ and $Y$ is $N(np,np(1-p))$, then $P(X=m)$ is still well-approximated by, for example, $P(m \leq Y \leq m+1)$. It is just that $P(m-1/2 \leq Y \leq m+1/2)$ is a better approximation, especially if $m$ is close to $np$.
Best Answer
Let $X$ be the random number of correct answers out of $n = 60$ true/false questions. The probability a single question is answered correctly is $p = 0.5$. Thus the exact distribution of $X$ is binomial: $$ X \sim {\rm Binomial}(n = 60, p = 0.5).$$ In order to get a passing grade of at least $70\%$, the required number of correct responses is $X \ge (0.7)(60) = 42$. Therefore, the desired probability is $$\Pr[X \ge 42] = \sum_{k=42}^{60} \binom{60}{k} (0.5)^k (1 - 0.5)^{60-k}.$$ This sum is tedious to compute (there are $60 - 42 + 1 = 19$ terms), so we use the normal approximation to the binomial distribution with continuity correction: if $np \ge 10$ and $n(1-p) \ge 10$ (which are satisfied in this case), then $$X \overset{\bullet}{\sim} {\rm Normal}\left(\mu = np, \sigma = \sqrt{np(1-p)}\right),$$ approximately. Hence to calculate $\Pr[X \ge 42]$, we compare the exact distribution of $X$ to a normal distribution with mean $\mu = 30$ and standard deviation $\sigma = 3.87298$. With continuity correction, in order to account for the entire discrete probability mass at $X = 42$, we must write $$\Pr[X \ge 42] \approx \Pr[X > 41.5] = \Pr\left[\frac{X - \mu}{\sigma} > \frac{41.5 - 30}{3.87298}\right] = \Pr[Z > 2.96929],$$ where $Z$ is a standard normal random variable. This results in a probability of $0.00149246$, very small. The exact value using the binomial distribution is $$\tfrac{1539401708565319}{1152921504606846976} \approx 0.00133522.$$ This probability is small because the standard deviation is small: for so many questions, it is extremely unlikely that one could answer much more than half correctly by random chance.
If you want to do these computations in a TI calculator, then the normal approximation is given by the command
normalcdf(41.5,1E99,30,3.87298). The exact binomial probability is given by the command1-binomcdf(60,0.5,41)because the complement of the event that $X$ is at least $42$ is that $X$ is at most $41$.