Infinite product of fields

You are absolutely right when you write "I believe that the answer to (2) is that all primes are maximal", Pierre-Yves :

Proposition 1 The ring $A$ is zero-dimensional.
Proof Recall that a commutative ring $R$ is called von Neumann if for all $ r\in R$ there exists an $s\in R$ such that $r=sr^2$.
It is clear that a field is von Neumann and that a product of any family of von Neumann rings is von Neumann, so that your ring $A$ is von Neumann.
But a von Neumann ring has Krull dimension zero: indeed, if ${\mathfrak p } \subset A$ is prime, the quotient $D:=A/{\mathfrak p } $ is clearly von Neumann.
And to end the proof just note that a von Neumann domain $D$ is a field: if $d\neq 0\in D$, the relation $d=rd^2$ implies $1=rd $ so that $d$ is invertible.

Corollary For every prime (=maximal) ideal $\mathfrak p\subset A$ the residue field is $\kappa (\mathfrak p)=A_{\mathfrak p}$
Proof Since the ring $A$ is reduced, so is its localization, the zero-dimensional (cf.Proposition) ring $A_{\mathfrak p}$.
But then ${\mathfrak p} A_{\mathfrak p}$ is the nilpotent radical of that reduced ring $A_{\mathfrak p}$ and so must be zero: ${\mathfrak p} A_{\mathfrak p}=0$ .
Hence we get the (rather unusual !) formula $\kappa (\mathfrak p)=A_{\mathfrak p}/ \mathfrak pA_{\mathfrak p}=A_{\mathfrak p}/0=A_{\mathfrak p} $ .

And your other hope is true too:
Proposition 2 Every primary ideal of $A$ is maximal.
Indeed , since the ring $A$ is von Neumann it is absolutely flat (Atiyah-Macdonald, Introduction to Commutative Algebra, Chapter 2, Exercise 27 ) and thus every primary ideal of $A$ is maximal. (Same reference, Chapter 4, Exercise 3)

For (1), there will be variety results depended on the fields.

（1）If $A=\prod_{i\in \mathbb{N}}\mathbb{F}_p$ then all the residue fields are equal to $\mathbb{F}_p$. (since $x^p=x$)

(2) If $A=\prod_{p}\mathbb{F}_p$, then for any prime $\mathfrak{m}$ containing $\oplus_p\mathbb{F}_p$ , then $\mathbb{Q}\subseteq A/\mathfrak{m}$ and the cardinality of $A/\mathfrak{m}$ is $\aleph$, thus $A/\mathfrak{m}$ is not an algebraic extension. It has transcendental elements over $\mathbb{Q}$.

Proof about the cardinality: define a map $\varphi$ from $\{0,1\}^{\mathbb{N}}$ to $A/\mathfrak{m}$ by sending $(a_i)_{i=1}^{\infty}$ to $(x_k)$, $x_k=\sum_{j=1}^na_j2^{j-1}$ if $2^n-1<p_k\leq2^{n+1}-1$. Here index $k$ denotes $k$-th coordinate and $p_k$ denotes the $k$-th prime number. Then $\varphi$ is injective.

In this case, we want to ask if $A$ is a purely transcendental extension over $\mathbb{Q}$ ?

It is not true at least for some maximal ideal!

Consider the equation $x^2+1=0$. Then it can be solved in $\mathbb{F}_p$ if only if $p=1\mod 4$ or $p=2$. Since those primes are infinite, let $a=(a_{p_k})\in A$, $a_{p_k}=1$ if $p_k=1 \mod 4$, then we find a maximal ideal $\mathfrak{m}$ containing $\oplus\mathbb{F}_p$ such that $1-a\in \mathfrak{m}$. Hence $x^2+1=0$ can be solvable in $A/\mathfrak{m}$.

But also we can find some maximal ideal $\mathfrak{m}$ such that the $x^2+1=0$ can not be solvable in $A/\mathfrak{m}$.

The residue fields are always not purely algebraic extension. Since $(x^2+1)(x^2+2)(x^2-2)$ can be always solvable.

(3) If $A=\prod_{i\in \mathbb{N}}\mathbb{Q}$, the discussion about the cardinality of the residue field is similar.

Those questions relate to my recent thread the ring of Cauchy sequence.

Another way to show that $\dim A=0$:

suppose we have a proper inclusion between prime ideals, say $\mathfrak{p}\subsetneq\mathfrak{m}$, then you can find an $x\in \mathfrak{m}-\mathfrak{p}$ such that the coordinate of $x$ is either $1$ or $0$, then we must have $1-x\in\mathfrak{p}\subset \mathfrak{m}$, contradiction!