Picking zero out of zero elements

First, how many ways are there to choose a subcommittee of $4$ Republicans, $3$ Democrats, and $1$ Libertarian out of a committee of $8$ Republicans, $6$ Democrats, and $2$ Libertarians? The answer is $$ \binom 8 4 \binom 6 3 \binom 2 1. $$ So how many ways are there to choose a subcommittee of $4$ Republicans, $3$ Democrats, and $0$ Libertarians out of a committee of $8$ Republicans, $6$ Democrats, and $0$ Libertarians? This should be $$ \binom 8 4 \binom 6 3 \binom 0 0. $$ But it should also be $$ \binom 8 4 \binom 6 3, $$ since you can just omit all mention of Libertarians in that case.

The number of $0$-element subsets of $0$-element set is $1$. Indeed, the only empty subset of $\emptyset$ is $\emptyset$. :-)

There is only one way to choose $0$ objects from any sized set - To not choose. That makes as much sense if the set itself is empty as if the set has multiple objects.

Note also that $0! = 1$, for good reasons: it's convenient to define the empty product to be the identity, because then we get $$\prod_{x \in A} x \times \prod_{y \in B} y = \prod_{z \in A \cup B} z$$ for disjoint $A, B$ even when $A = \emptyset$.

Other answers correctly address the first part of your question. For the second, the number of ways to choose one blue candy is $\binom{0}{1} = 0$: the number of ways to choose one item from an empty set. You don't evaluate that with the formula for the binomial coefficients. Use it for the numerator and you get the correct $0$ answer.

By definition, $0! = 1$. Think of it like this: how many ways can we arrange no objects? Easy - one way, the way that arranges no objects.

Definition 1.1 of https://www.math.ucdavis.edu/~hunter/intro_analysis_pdf/ch1.pdf is helpful to understand this.

Consider the power set $\mathcal{P}(X)$ of a set $X$, which is the set of all subsets of $X$. Note that this includes the empty set $\emptyset$. Picking any number (including 0) of items from $X$ is equivalent to picking one of the items in $\mathcal{P}(X)$. Thus, the number of ways of picking any number $n$ of items from $X$ is equal to the number of occurrences in $\mathcal{P}(X)$ of a set of length $n$.

If we select $n=0$, there is only one set of length $0$, namely the empty set $\emptyset$ which exists only once in $\mathcal{P}(X)$. Thus follows ${m \choose 0}=1$ for any $m \in \mathbb{N}$.