We are tossing $20$ dice. Let $X_i$ be the number showing on the $i$-th die. We are interested in the random variable $X_1+\cdots+X_{20}$.
The $X_i$ are independent, mean $\frac{7}{2}$, variance $\frac{35}{12}$ (please verify).
So $Y$ has mean $\mu=20\cdot\dfrac{7}{2}$, variance $\sigma^2=20\cdot \dfrac{35}{12}$.
We cross our fingers and use the normal approximation. This is because $Y$ is the sum of a not terribly small number of independent identically distributed respectable random variables $X_i$.
So we want the probability that if $W$ is normal with mean $\mu$ and variance $\sigma^2$, then $30\le W\le 40$.
The rest depends to some degree on whether you are expected to use the continuity correction.
Without the continuity correction, we want the probability that a standard normal $Z$ satisfies $\frac{30-\mu}{\sigma}\le Z\le \frac{40-\mu}{\sigma}$.
With continuity correction, replace $40$ by $40.5$, and $35$ by $34.5$.
If you have any difficulty finishing, please leave a comment.
(SIde note: The events are neither mutually exclusive nor independent.)
$P(E\cap F)$ is the probability that one die is a six and the other die is not. That's $\frac 1 6\frac 5 6+\frac 5 6\frac 1 6$ by adding the probability that the first die is a six and the other not, to the probability that the first die is not a six and the other is. (NB: Those events are mutually exclusive partitions of $E\cap F$.)
$$\mathsf P(E\cap F) = 2 \cdot \frac 1 6 \cdot \frac 5 6\\ = \frac{10}{36}$$
Then we just use conditional probability as you noted.
$$\mathsf P(E\mid F) = \frac{\mathsf P(E\cap F)}{\mathsf P(F)}\\ = \frac{10/36}{30/36} \\ = \frac {1}{3}$$
Best Answer
$P(C)$ is actually $\frac{11}{36}$ – $11$ of the $36$ possible rolls show at least one 1 (don't forget to consider the double-1 case!). $P(A\cap C)=\frac2{36}$ since only 1-2 and 2-1 have at least one 1 and sum to $3$. Thus $$P(A|C)=\frac{P(A\cap C)}{P(C)}=\frac{2/36}{11/36}=\frac2{11}$$