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
$$\binom{9}{3}p^3(1-p)^6.$$
Best Answer
If $X$ is normally distributed with mean $\mu$ and standard deviation $\sigma$, then the average of $n$ independent samples from $X$ is normally distributed with mean $\mu$ and standard deviation $\dfrac{\sigma}{\sqrt n}$.
In your case $\sigma=2$, $n=16$.