1)
Let $A_i = k[x_i]$.
There exists a canonical injection $A_i/m_i \rightarrow A/m$.
Since $A/m$ is finite algbraic over $k$, so is $A_i/m_i$.
Hence $m_i$ is maximal.
2)
Let $Specm(A)$ be the set of maximal ideals of $A$.
Let $\bar k$ be an algebracic closure of $k$.
Let $Homalg_k(A, \bar k)$ be the set of $k$-homomorphisms.
Let $f \in Homalg_k(A, \bar k)$.
Then $Ker(f) \in Specm(A)$.
Conversely let $m \in Specm(A)$.
It is easy to see that there exists $f \in Homalg_k(A, \bar k)$ such that $m = Ker(f)$.
Let $f, g \in Homalg_k(A, \bar k)$.
We write $f \equiv g$ if there exists $\sigma \in Aut(\bar k/k)$
such that $\sigma f = g$.
This is an equivalent relation on $Homalg_k(A, \bar k)$.
Let $f, g \in Homalg_k(A, \bar k)$.
Let $m = Ker(f), m' = Ker(g)$.
Then $m = m'$ if and only if $f \equiv g$.
Hence $Specm(A)$ is identified with the set of equivalence classes on $Homalg_k(A, \bar k)$
Similar results hold for $A_i$.
Paraphrasing above results we get the following criteria:
Let $f, g \in Homalg_k(A, \bar k)$.
Let $m = Ker(f), m' = Ker(g)$.
Let $\alpha_i = f(x_i), \beta_i = g(x_i)$ for all $i$.
Then $m = m'$ if and only if there exists $\sigma \in Aut(\bar k/k)$ such that $\sigma(\alpha_i) =\beta_i$ for all $i$.
For each $i$, $m_i = m'_i$ if and only if there exists $\tau_i \in Aut(\bar k/k)$ such that $\tau_i(\alpha_i) =\beta_i$.
Let $m_i \in Specm(A_i)$ for $i = 1, \dots, n$.
There exists a monic polynomial $f_i(x_i) \in A_i$ such that $m_i = (f_i)$ for each $i$.
Let $P_i$ be the set of roots of $f_i(x_i)$ in $\bar k$.
Let $P = \prod_i P_i$.
Let $G = Aut(\bar k/k)$.
$G$ acts on $P$ in the obvious way.
Let $P/G$ be the set of $G$-orbits.
Let $M$ be the set of maximal ideals of $A$ lying over $m_i$ for all $i$.
Let $\alpha = (\alpha_1,\dots,\alpha_n) \in P$.
There exists a unique $f \in Homalg_k(A, \bar k)$ such that $f(x_i) = \alpha_i$ for all $i$.
We denote by $\psi(\alpha) = Ker(f)$.
Clearly $Ker(f) \in M$.
Hence we get a map $\psi\colon P \rightarrow M$.
By the above result, $\psi$ induces a bijection $\bar \psi\colon P/G \rightarrow M$.
Hence $|P/G| = 1$ if and only if there exists only one maximal ideal of $A$ lying over $m_i$ for all $i$.
Now we construct an example such that $m \neq m'$ and $m_i = m'_i$ for all $i$.
Let $\alpha, \beta \in \bar k$ be separable over $k$.
Suppose $k(\beta) \supset k(\alpha)$
and $[k(\beta) : k(\alpha)] > 1$ and $[k(\alpha) : k] > 1$.
Let $p(x)$ be the minimal polynomial of $\beta$ over $k$.
Let $q(x)$ be the minimal polynomial of $\beta$ over $k(\alpha)$.
Clearly deg $p(x) >$ deg $q(x)$.
Since $p(x)$ is divisible by $q(x)$, there exists a root $\beta'$ of $p(x)$ such that $\beta'$ is not a root of $q(x)$.
Let $A = k[x_1, x_2]$.
Let $f\colon A \rightarrow \bar k$ be the map defined by $f(x_1) = \alpha$, $f(x_2) = \beta$.
Let $g\colon A \rightarrow \bar k$ be the map defined by $g(x_1) = \alpha$, $g(x_2) = \beta'$.
Let $m = Ker(f), m' = Ker(g)$.
Suppose $m = m'$.
There exists an automorphism $\sigma$ of $\bar k/k$ such that $\sigma(\alpha) = \alpha$, $\sigma(\beta) = \beta'$.
Hence $\beta'$ is a root of $q(x)$.
This is a contradiction.
Hence $m \neq m'$.
Clearly $m_1 = m'_1$ and $m_2 = m'_2$.
