How do you find $f(x_1, x_3)$? $X_i$ is the number of times (out of 100) that a die's face has $i$ dots.
I know that $X_i\sim \text{binomial}(100, 1/6)$, so $f(x_i)={100 \choose x_i}(1/6)^{x_i}(5/6)^{100-x_i}$.
How do you find the joint probability mass function for $X_1$ and $X_3$ (the number of times you roll $1$s and $3$s, respectively? I'm unfamiliar with joint pmfs so I'm not sure how to even begin figuring this out.

One way is to figure it out from the conditional PMF instead. First, a priori the distribution of $X_1$ is binomial($100$,$1/6$), as you've said. Now suppose you know the value of $X_1$ is $y$. Then you have $100-y$ remaining chances to roll a $3$. The only dependence between these two groups of rolls is that you can no longer roll a $1$ in the second group. Hence the conditional PMF of $X_3$ given $X_1$ is the PMF for binomial($100-X_1$,$1/5$).
Then we have the formula
$$p_{X_1,X_3}(x,y) = p_{X_3|X_1}(x,y) p_{X_1}(y).$$
Looking at the right side, the first term is, as a function of $x$ for each fixed $y$, the PMF of binomial($100-y$,$1/5$). The second term is the PMF of binomial($100$,$1/6$). Can you put it together from here?
