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Here is a simple argument for the second question. I renamed the field $k$ to $K$ for clarity.

Observe that $K/\mathbb{Q}$ is a Galois extension, K^{\mathrm{Aut}(K/\mathbb{Q})}=\mathbb{Q}$, because $K$ is algebraically closed. For any $g\in\mathrm{Gal}(K/\mathbb{Q})$ g\in\mathrm{Aut}(K/\mathbb{Q})$ and any $x\in X$ we have $x^g\in X$, since $$ P_l(x^g) = P_l^g(x^g) = P_l(x)^g = 0^g=0,\qquad 1\leq l\leq k,$$ using that the coefficients of $P_l$ are rational. This shows that the polynomials $$ Q_j(z):=\prod_{i=1}^m(z-z_{ij}),\qquad 1\leq j\leq n, $$ are invariant under $\mathrm{Gal}(K/\mathbb{Q})$, \mathrm{Aut}(K/\mathbb{Q})$, i.e. they also have rational coefficients. It follows that the coordinates $z_{ij}\in K$ are algebraic over $\mathbb{Q}$.

P.S. Thanks to Qiaochu Yuan and Kevin Ventullo.
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Here is a simple argument for the second question. I renamed the field $k$ to $K$ for clarity.

Observe that $K/\mathbb{Q}$ is a Galois extension, because $K$ is algebraically closed. For any $g\in\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ g\in\mathrm{Gal}(K/\mathbb{Q})$ and any $x\in X$ we have $x^g\in X$, because since $$ P_l(x^g) = P_l^g(x^g) = P_l(x)^g = 0^g=0,\qquad 1\leq l\leq k,$$ using that the coefficients of $P_l$ are rational. This shows that the polynomials $$ Q_j(z):=\prod_{i=1}^m(z-z_{ij}),\qquad 1\leq j\leq n, $$ are invariant under $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$, \mathrm{Gal}(K/\mathbb{Q})$, i.e. they also have rational coefficients. It follows that the points coordinates $z_{ij}$ z_{ij}\in K$ are algebraic .over $\mathbb{Q}$.
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Here is a simple argument for 2): Every element of $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ fixes the polynomials second question.

For any $P_l$, hence g\in\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ and any $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ permutes the finite set x\in X$ we have $X$. It follows that for each x^g\in X$, because $$ P_l(x^g) = P_l^g(x^g) = P_l(x)^g = 0^g=0,\qquad 1\leq j\leq n$ l\leq k,$$ using that the polynomial coefficients of $Q_j(z):=\prod_{i=1}^m(z-z_{ij})$ has P_l$ are rationalcoefficients. This shows that the polynomials $$ Q_j(z):=\prod_{i=1}^m(z-z_{ij}),\qquad 1\leq j\leq n, because it is $$ are invariant under $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$. This shows \mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$, i.e. they have rational coefficients. It follows that each the points $z_{ij}$ is are algebraic.
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