Let $Y_1, Y_2, \dots,Y_9$ be the IQ measurements. Then
$$\bar{Y}=\frac{Y_1+Y_2+\cdots+Y_9}{9}.$$
It is a standard fact that I imagine you know that under your assumptions, $\bar{Y}$ has normal distribution, with mean the mean of the $Y_i$, and standard deviation $\frac{\sigma}{\sqrt{9}}$, where $\sigma$ is the standard deviation of the $Y_i$.
Thus in our case, the standard deviation of $\bar{Y}$ is $\frac{16}{3}$.
Now we tackle the problem of the probability that $\bar{Y}\gt 103$. Here there is some ambiguity, because published IQ's are usually integers. But we will assume they can be in principle any real number. Then
$$\Pr(\bar{Y}\gt 103)=\Pr\left(Z\gt \frac{103-100}{16/3}\right),$$
where $Z$ is standard normal. You can now obtain the probability from a table of the standard normal, or software. I think the answer is about $0.2868$.
For the second problem and third problem, let $p$ be the probability that an individual's IQ exceeds $103$. This is the probability that a standard normal is $\gt \frac{103-100}{16}$, and can be found using a table or software.
I get about $0.4257$. This is close to your answer, so I am sure you did it more or less the right way. I am almost sure that you rounded $\frac{3}{16}$ to $0.18$. It is actually $0.1875$, so three-quarters of the way to $0.19$.
Call exceeding $103$ a success. We want the probability of exactly $3$ successes in $9$ trials. For the answer, we use the Binomial distribution. The required probability is
$$\binom{9}{3}p^3(1-p)^6.$$
We will assume that by "difference does not exceed $0.5$ inch," what is meant is that the absolute value of the difference does not exceed $0.5$ inch.
Let the heights of the people in the sample be $X_1,X_2,\dots,X_{100}$. Then the sample mean is the random variable $Y$, where
$$Y=\frac{1}{100}(X_1+X_2+\cdots+X_{100}).$$
Let the population mean be $mu$. Then $Y$ has mean $\mu$ and standard deviation $\frac{2.5}{\sqrt{100}}$.
We assume that the distribution of the $X_i$ is reasonably nice. Then $Y$ has a close to normal distribution. From here on, we assume the distribution is normal. Then $Y-\mu$ is normal with mean $0$ and standard deviation $0.25$.
Then
$$\Pr\left(|Y-\mu|\le 0.5\right)=\Pr\left(|Z|\le \frac{0.5}{0.25}\right).$$
This can be found using tables of the standard normal, or software.
Best Answer
This is not negative.
$$\Phi(1) + \Phi(0.5)-1=0.5318$$