$$P(D_2=R) = P(D_2=R|D_1=B)\times P(D_1=B) + P(D_2=R|D_1=R)\times P(D_1=R)$$
$$P(D_2=R)=\frac{5}{9} \times \frac{3}{8}+\frac{6}{9} \times \frac{5}{8}= \frac{5}{8}$$
Now,
$$P(D_1=B|D_2=R) = \frac{P(D_1=B,D_2=R)}{P(D_2=R)} = \frac{\frac{3}{8}\frac{5}{9}}{\frac{5}{8}} = \frac{1}{3}$$
Does 3 in $X_i\sim Binomial(3,\theta)$ mean there are 3 balls in the bag?
No. If you want to apply Example 8.8 to the 'balls in the bag' example that the tutorial gives earlier, the 3 means that you do the following action three times: draw a ball, record if its blue, and put it back in the bag (this last part is very important). The total number of blue balls you record is $X_i$ and is a binomial random variable. It will take on values of 0, 1, 2, or 3. You could do this if there were just one ball in the bag or if there were 100 balls in the bag.
In example 8.8, it is telling you that the entire process in the paragraph above was repeated 4 times, and hence you have a set of 4 numbers, each of which could be 0, 1, 2, or 3.
It was perhaps unwise of the tutorial author to use $\theta$ for both the $Bernoulli(\theta/3)$ example (where $\theta$ is the number of blue balls in the bag) and the $Binomial(3,\theta)$ example, where $\theta$ represents the probability of success (i.e. drawing one ball and getting a blue ball).
=====Addition in response to OP's edits=====
A Binomial with $n$ trials is defined as the sum of $n$ independent Bernoulli random variables having a common probability of success, and the sum of independent Binomial random variables each having a common probability of success is itself a Binomial random variable with number of trials equal to the sum of the constituent number of trials. Thus, if $\mu$ is a probability of success (I'm changing the parameter from $\theta$ to $\mu$ so that it is clear I'm not using the tutorial's parametrization), all of the following experiments yield equivalent statistical information (and have the same likelihood up to a multiplicative constant) about $\mu$:
(i) 20 independent draws from $Bernoulli(\mu)\equiv Binomial(1,\mu)$
(ii) 10 independent draws from $Binomial(2,\mu)$
(iii) 5 independent draws from $Binomial(4,\mu)$
(iv) 1 draw from $Binomial(20,\mu)$
The parameter $\mu$ has the same interpretation in all cases: the probability of success for a single Bernoulli random variable.
Best Answer
Let $B$ denotes blue balls, $R$ denotes red balls, then you may apply the formula for hypergeometric distribution:
$$P(B = 20, R = 0) = \frac{\binom{10}{0}\binom{90}{20}}{\binom{100}{20}} = \frac{\binom{90}{20}}{\binom{100}{20}}$$
The last term exactly matches the @Macro's answer, but hypergeometric formula is more general. The idea beyond the formula is simple: get the number of ways to draw $20$ $B$ of $90$, number of ways to draw $0$ $R$ from $10$ (there is only one possibility) and divide their product by the number or ways to draw any $20$ balls from $100$. Hope this was not your homework ;)