Question 1
I'm not sure what you mean by $Y$ here, but suppose you have a family of topological space $(X_\alpha)_{\alpha \in A}$, and we define the product $Y = \prod_{\alpha \in A} X_\alpha$ (our index set $A$ need not be countable). The elements of $Y$ are essentially functions $f$, mapping from $A$ to the union of all the $X_\alpha$s, where $f(\alpha) \in X_\alpha$. For each $\alpha \in A$, we can naturally define a projection map $\pi_\alpha$, which takes $f \in Y$, and maps it to $f(\alpha)$, giving us a map from $Y$ to $X_\alpha$.
(In the case of a finite product, we can make our index set $A = \{1, 2, \ldots, n\}$ for some $n$. We can also think of our elements of $Y$ as $n$-tuples. Then, given an $n$-tuple $y \in Y$, $\pi_i(y)$ is the $i$th coordinate of $y$. Outside of finite/countable settings, it makes more sense to think of the elements as functions rather than tuples.)
If $U_\alpha$ is open in $X_\alpha$, the set $\pi_{\alpha}^{-1}(U_\alpha)$ is the set of functions in $Y$ satisfying $f(\alpha) \in U_\alpha$. For other indices $\beta \neq \alpha$, $f(\beta)$ is unrestricted, other than the usual restriction of $f(\beta) \in X_\beta$. So we have
$$\pi^{-1}_\alpha(U_\alpha) = \prod_{\beta \in A} X'_\beta,$$
where
$$X'_\beta = \begin{cases} X_\beta & \text{if } \beta \neq \alpha \\ U_\alpha & \text{if } \beta = \alpha \end{cases}.$$
Question 2
This is a subbasis because it is not necessarily the case that the intersection of two of these sets will be a union of basis elements. Even if we restrict ourselves to $\mathbb{R} \times \mathbb{R}$, these sets will not be a basis. For example, we can express an open square in the form:
$$(0, 1) \times (0, 1) = ((0, 1) \times \mathbb{R}) \cap (\mathbb{R} \times (0, 1)) = \pi_1^{-1}(0, 1) \cap \pi_2^{-1}(0, 1),$$
making it the intersection of two (presumably) open sets in our (supposed) topology, which should make it open. But, this is a bounded set, and $\pi_i^{-1}(U)$ is unbounded for any open $U \subseteq \mathbb{R}$ and $i \in \{0, 1\}$. So, there's no way to express the square as a union of our supposedly basic sets, which makes the sets not basic!
To form a basis from a subbasis, all you do is take all finite intersections of your subbasic sets. So, our basis will be sets of the of the form
$$\pi_{\alpha_1}^{-1}(U_{\alpha_1}) \cap \ldots \cap \pi_{\alpha_n}^{-1}(U_{\alpha_n}),$$
where $\alpha_1, \ldots, \alpha_n \in A$ and $U_{\alpha_i} \in X_{\alpha_i}$ for all $i$. Equivalently, they are sets of the form
$$\prod_{\alpha \in A} X'_\alpha,$$
where $X'_\alpha = X_\alpha$ for all $\alpha \in A$ except for a number of $\alpha$s. For such $\alpha$s, $X'_\alpha$ is an open subset of $X_\alpha$.
Question 3
The box topology produces sets of the form
$$\prod_{\alpha \in A} X'_\alpha,$$
where $X'_\alpha$ is an open subset of $X_\alpha$ for all $\alpha \in A$. That is, we don't require $X'_\alpha = X_\alpha$ for all but a finite number of $\alpha$s. Every coordinate can be restricted at the same time.
For example, take the product $Y = \prod_{i=1}^\infty \mathbb{R}$, and the subset $U = \prod_{i=1}^\infty (-i, i)$. Since $(-i, i) \subseteq \mathbb{R}$ is open for all $i$, $U$ is open in $Y$ under the box topology. But, $(-i, i) \neq \mathbb{R}$ for infinitely many $i$ (in fact, for all $i$), so $U$ is not a basic element in the product topology.
This doesn't necessarily mean it's not open in the product topology, but this is true as well. If it were open in the product topology, it would have to contain a non-empty basic set. Such a set would have to take the form
$$U_1 \times U_2 \times \ldots \times U_n \times \mathbb{R} \times \mathbb{R} \times \ldots,$$
where $U_1, \ldots, U_n$ are open subsets of $\mathbb{R}$ (possibly equal to $\mathbb{R}$). But, this means that the $n + 1$th coordinate is unrestricted, but yet is still confined to $(-n - 1, n + 1)$, which is a contradiction. So, no basic set can be contained in $U$, and hence $U$ is not open with respect to the product topology.
Best Answer
Note first that your post (in my version of Munkres this is correct) contains a typo. $\mathcal{S}$ is a subbasis of $X \times Y$ and not of $X$.
In your case: $$\{\pi_1^{-1}(U): U \in \tau_X\}= \{\pi_1^{-1}(\emptyset),\pi_1^{-1}(\{1\}), \pi_1^{-1}(\{1,2\})\}= \{\emptyset, \{1\}\times Y, X \times Y\}$$ Similarly
$$\{\pi_2^{-1}(U): U \in \tau_X\}= \{\pi_2^{-1}(\emptyset),\pi_2^{-1}(\{1\}), \pi_2^{-1}(\{1,2\})\}= \{\emptyset, X\times \{3\}, X \times Y\}$$
Thus, $$\mathcal{S}= \{\emptyset, X \times Y, \{1\}\times Y, X\times \{3\}\}$$ $$=\{\emptyset, \{(1,3),(1,4),(2,3),(2,4)\}, \{(1,3),(1,4)\}, \{(1,3),(2,3)\}\}$$ is a subbasis of $X \times Y$.
It is not a basis, because a basis must contain $\{(1,3)\}= \{1\}\times \{3\}$.