We can show that the answer to Q2 is "No" even without appealing to the Central Limit Theorem.
Let's say $H$ is the total number of heads. By the Chebyshev inequality (see below),
$$P(|H-70| \ge 21) \le \frac{21}{21^2} \approx 0.048$$
But
$$P(|H-70| \ge 21) = P(H \le 49) + P(H \ge 91)$$
so
$$P(H \le 49) \le P(|H-70| \ge 21) \le 0.048$$
Chebyshev's inequality: If $X$ is a random variable with finite mean $\mu$ and variance $\sigma^2$, then for any value $k>0$,
$$P(|X-\mu| \ge k) \le \frac{\sigma^2}{k^2}$$
I will show you the general method for such a problem
and let you plug in the specific numbers for your particular problem.
Suppose bricks have weights distributed as
$\mathsf{Norm}(\mu, \sigma).$ [Notice that in this notation the second parameter is the standard deviation, not the variance.]
(a) The mean $Y = \bar X$ of $n = 9$ such bricks
is distributed as $\bar X \sim \mathsf{Norm}(\mu, \sigma/\sqrt{n}).$ [That is, $SD(\bar X) = \sigma/\sqrt{n}.]$
(b) You seek
$$P(\bar X > a) =
P\left(\frac{\bar X - \mu}{\sigma/\sqrt{n}} >
\frac{a - \mu}{\sigma/\sqrt{n}}\right)
= P\left(Z > \frac{a - \mu}{\sigma/\sqrt{n}}\right),$$
where $Z$ has a standard normal distribution.
In the final expression, let $a = 1, \mu=0.96, \sigma = 0.045, n = 9.$ and use a printed table of the standard
normal cumulative distribution function to get your answer.
Using R, in which $\mathtt{pnorm}$ is a normal CDF with
specified parameters, one can get the numerical answer $P(\bar X > 1) = 0.00383$ as shown below. (Because of rounding, your result using printed normal tables will be
slightly less precise.)
n = 9; mu = 0.96; sg = 0.045
1 - pnorm(1, mu, sg/sqrt(n)) # last param is SD
[1] 0.003830381
In the following picture, the dotted density curve is for the
population of bricks, the solid curve is for the mean of $n = 1$ bricks, and the (small) area under the solid curve to the right of the vertical line represents the desired probability.
Notes: (1) Roughly speaking the PDF of the population is three times
as wide as the PDF for the PDF of the mean of nine. Thus
in order to include total probability 1 under the curve,
the PDF of the mean must be three times as tall.
(2) R is excellent statistical software, available free of charge for Windows, Mac and Unix computers from here. It is easy to learn to use--provided you use just what you need at each step.
Best Answer
All three questions can be solved in a pretty similar fashion. The key to answering them is using the rule that, if $A$ and $B$ are independent events, then $P(A\cap B) = P(A)P(B)$ — in other words, multiply the probabilities of each event. So for example, the probability of selecting a parasite shorter than 20 micrometers is $0.01 + 0.34 = 0.35$. The probability of selecting two such parasites, then, is $(0.35)(0.35) = 0.1225$.
There is a twist in the final question. The second question dictates what order you select the parasites in. But the third question states simply that you pick two parasites. Therefore, you could pick the shorter parasite followed by the longer one or the longer one followed by the shorter one. If $A$ is the first event and $B$ the second, you're seeking $P(A\cup B)$. If $P(A\cap B) = 0$ — which is the case here; it's impossible to pick both the shorter parasite and the longer one first — then $P(A\cap B) = P(A) + P(B)$.
Does that make sense?