Ellie rolls a normal, six-sided die and writes down the scores. She rolls it until she has seen three different scores. What is the probability that
she needs more than 3 rolls?
P(rolls needed > 3) = 1 - P(rolls needed = 3)
P(rolls needed = 3) = P(2nd roll $\not =$ 1st roll) $\cdot$ P(3rd roll $\not =$ 1st and 2nd rolls|1st roll $\not =$2nd roll) = $\frac{5}{6}\frac{4}{6} = \frac{5}{9}$
So P(rolls needed > 3) $= 1 - \frac{5}{9} = \frac{4}{9}$
the three distinct scores add up to an odd number, given that her first roll was a 1?
As the first roll is 1, the second two rolls must add up to an even number. So the answer is $\frac{\text{number of ways to have two unique numbers between 2 and 6 add up to an even}}{\text{number of ways to have two unique numbers between 2 and 6}}$. To add up to an even, the second two rolls must be either both odds, or both evens. How many unique pairs of evens exist? 3 (either 2 and 4, 2 and 6, or 4 and 6). How many unique pairs of odds exist? 1 (3 and 5). So there are 4 possible pairs of unique rolls that would meet this requirement. But how many possible unique pairs are there total? 10. The 4 already mentioned, plus the 6 pairs of an odd and an even (2 and 3, 2 and 5, 3 and 4, 3 and 6, 4 and 5, 5 and 6). You can use this to find the answer.
the three distinct scores are consecutive numbers (ignoring the order in which they appeared)?
How many distinct ways can 3 numbers between 1 and 6 be consecutive? You can write them all down: (1,2,3), (2,3,4), etc. That will be the numerator for your probability fraction. The denominator will be the number of ways to choose 3 distinct integers between 1 and 6, $ \binom{6}{3}$. Are you familiar with combinations?
Let's call the rolled dice $A,B,C$
she needs more than 3 rolls?
The probability for needing exactly three rolls is: First one can be any; second one can be 5 out of 6 due to as $A \neq B$, third one is 4 out of 6 for $C \neq A, C \neq B$. That is $Pr = 1 \cdot \frac{5}{6} \cdot \frac{4}{6} = \frac{5}{9}$. The probability for needing more than three rolls is $1 - \frac{5}{9} = \frac{4}{9}$
the three distinct scores add up to an odd number, given that her first roll was a 1?
Assume that $A = 1$. Then $B$ is either even or odd, both with probability $1/2$. And again $C$ is even or odd with both probabilities at $1/2$. If both are even or both are odd, the sum $B+C$ is even, if one of them is odd and the other even, the sum $B+C$ is odd, meaning the sum is also distributed uniformly ($1/2$ each). For the overall sum this means it is also distributed uniformly with $1/2$.
the three distinct scores are consecutive numbers (ignoring the order in which they appeared)?
Valid triples: $(1,2,3),(2,3,4),(3,4,5),(4,5,6)$, out of $6^3 = 216$ total combinations for three dice (with respect to the order). For each of those triples, there are $3! = 6$ possible orders to draw them (e.g. $(1,2,3),(1,3,2),(2,1,3),(2,3,1),(3,1,2),(3,2,1)$). That means in total, $4 \cdot 6 = 24$ of the combinations (with order) out of the $216$ total combinations are valid. That results in the probability $\frac{24}{216} = \frac{1}{9}$.