The answer to (a) is clearly $\dfrac{7}{20}$. This is because every one of the $20$ balls is equally likely to be the third ball drawn.
For (b), let $R$ be the event that the third ball drawn is red. Let $B$ be the event the first two balls drawn are blue. We want the conditional probability $\Pr(B|R)$. By the usual expression for conditional probability, we have
$$\Pr(B|R)=\frac{\Pr(B\cap R)}{\Pr(R)}.$$
We already know $\Pr(R)$, so all we need is $\Pr(B\cap R)$. This is not hard to calculate:
$$\Pr(B\cap R)=\frac{13}{20}\cdot\frac{12}{19}\cdot \frac{7}{18}.$$
$1.$ With replacement: There are three ways this can happen, red, red, red; blue, blue, blue; and green, green, green.
The probability of red, red, red is $\left(\dfrac{5}{19}\right)^3$. Find similar expressions for the other two colurs, and add up.
$2.$ Without Replacement: The probability of red, red, red is $\left(\dfrac{5}{19}\right)\left(\dfrac{4}{18}\right)\left(\dfrac{3}{17}\right)$. Find similar expressions for the other two colours, and add up.
Remark: We could also do the problem by a counting argument. Let's look at the first problem. Put labels on the balls to make them distinct. Then for with replacement, there are $19^3$ strings of length $3$ of our objects. (The three objects in the string are not necessarily distinct.) All of these $19^3$ strings are equally likely.
There are $5^3$ all red strings, $11^3$ all blue, and $8^3$ all green. This gives probability $\dfrac{5^3+6^3+8^3}{19^3}$.
One can do a similar calculation for the without replacement case.
For without replacement, there is a third approach. We can choose $3$ objects from $19$ in $\dbinom{19}{3}$ ways. And we can choose $3$ of the same colour in $\dbinom{5}{3}+\dbinom{6}{3}+\dbinom{8}{3}$ ways.
Best Answer
$$P(RRXX)=\frac{9}{21}\times \frac{8}{20}\times \frac{19}{19} \times \frac{18}{18}=\frac{6}{35}$$
Where $X$ indicate: any color