It seems to me that the answer on both questions is Yes.

Let $F$ be a finitely generated free non-abelian pro-$p$ group, $N=\overline{F^{\prime\prime}}$ and $K=\overline {[N,F]}$. Then $N/K$ is torsion free and of infinite rank. Let $a_1,a_2,\ldots$ be a free $\mathbb Z_p$-generating set of $N/K$. Put $G_1=F/\langle K, a_i^{p^i}\rangle$. Let $a=\prod a_i$ and let $\bar a$ be its image in $G_1$. Then $\bar a$ is not a torsion element but lies in the clousure of the subgroup of torsion elements. This answers the first question.

Let $G_2$ be the quotient of $G_1\times \langle z\rangle $ by the subgroup generated by $z^p\bar a$. Then the image of $z$ in $G_2$ is not a torsion element.

It seems to me that the answer on both questions is Yes.

Let $F$ be a finitely generated free non-abelian pro-$p$ group, $N=\overline{F^{\prime\prime}}$ and $K=\overline {[N,F]}$. Then $N/K$ is torsion free and of infinite rank. Let $a_1,a_2,\ldots$ be a free $\mathbb Z_p$-generating set of $N/K$. Put $G_1=F/\langle K, a_1^{p},a_2^{p^2},\ldots\rangle$. Let $a=\prod a_i$ and let $\bar a$ be its image in $G_1$. Then $\bar a$ is not a torsion element but lies in the clousure of the subgroup of torsion elements. This answers the first question.

Let $G_2$ be the quotient of $G_1\times \langle z\rangle $ by the subgroup generated by $z^p\bar a$. Then the image of $z$ in $G_2$ is not a torsion element.

It seems to me that the answer on both questions is Yes.

Let $F$ be a finitely generated free non-abelian pro-$p$ group, $N=\overline{F^{\prime\prime}}$ and $K=\overline {[N,F]}$. Then $N/K$ is torsion free and of infinite rank. Let $a_1,a_2,\ldots$ be a free $\mathbb Z_p$-generating set of $N/K$. Put $G_1=F/\langle K, a_i^{p^i}\rangle$. Let $a=\prod a_i$ and let $\bar a$ be its image in $G_1$. Then $\bar a$ is not a torsion element but lies in the clousure of the subgroup of torsion elements. This answers the first question.

Let $G_2$ be the quotient of $G_1\times \langle z\rangle $ by the subgroup generated by $z^pa$. Then the image of $z$ in $G_2$ is not a torsion element.

It seems to me that the answer on both questions is Yes.

Let $F$ be a finitely generated free non-abelian pro-$p$ group, $N=\overline{F^{\prime\prime}}$ and $K=\overline {[N,F]}$. Then $N/K$ is torsion free and of infinite rank. Let $a_1,a_2,\ldots$ be a free $\mathbb Z_p$-generating set of $N/K$. Put $G_1=F/\langle K, a_i^{p^i}\rangle$. Let $a=\prod a_i$ and let $\bar a$ be its image in $G_1$. Then $\bar a$ is not a torsion element but lies in the clousure of the subgroup of torsion elements. This answers the first question.

Let $G_2$ be the quotient of $G_1\times \langle z\rangle $ by the subgroup generated by $z^p\bar a$. Then the image of $z$ in $G_2$ is not a torsion element.