Question 1 is true for countable $\omega^\xi$ but it is at least consistent that this is false at $\omega_1.$ There might be a ZFC counterexample but the first thing that popped to mind is a Suslin tree. It is consistent with ZFC that there is a Suslin tree but it is also consistent that there aren't any.

To see that a Suslin tree $X$ is a counterexample, recall that a Suslin tree is a tree of height $\omega_1$ where every linearly ordered set is countable and every antichain is countable. Suppose we have a collection $(X_i)_{i \in I}$ of incomparable nonempty linearly ordered subsets of $X$. Let $x_i$ be the minimal element of $X_i$. This set $\{x_i \mid i \in I\}$ is an antichain, which must therefore be countable. Since each $X_i$ is countable, the supremum of the heights of the $X_i$ must be countable as well since $\omega_1$ is regular.

For Question 2, if $n \geq 2$, it is necessary that $\xi$ is either $\omega$ or a weakly compact cardinal otherwise the result is false when $X$ is linearly ordered. The result is true for $\xi=\omega$ because of Ramsey's Theorem. There are two cases:

First, the easy case, if $X$ has a branch of height $\omega$ then apply the infinite form of Ramsey's Theorem to that branch.

Otherwise, by Kőnig's Lemma, the tree must have an infinite level. For simplicity, let's chop the base of the tree so that we can assume that the tree (or rather forest) has infinitely many roots. This gives a partition of $X$ into infinitely many rooted trees $(X_i)_{i \in I}$. Since $X$ has height $\omega$, the heights of these trees are unbounded and they are all finite. From a selection of these trees pick branches $Y_1,Y_2,\ldots$ so that the finite form of Ramsey's Theorem ensures that if $[Y_m]^n$ is $k$-colored then there is a homogeneous set $Z_m$ of size $m$ (where $n$ and $k$ are given by your setup). Apply Ramsey's Theorem to each $Y_m$ using your coloring and assemble the subchains $Z_m$ into a homogeneous tree of height $\omega$.

The same trick applies when $\xi$ is weakly compact. The basic idea is that a weakly compact cardinal satisfies the analogue of the infinite form of Ramsey's Theorem and it also has an analogue of Kőnig's Lemma. As a replacement for the finite form of Ramsey's Theorem, one needs to use the Erdős-Rado Theorem along with the observation that weakly compact cardinals are inaccessible.

Question 1 is true for countable $\omega^\xi$ but it is at least consistent that this is false at $\omega_1.$ There might be a ZFC counterexample but the first thing that popped to mind is a Suslin tree. It is consistent with ZFC that there is a Suslin tree but it is also consistent that there aren't any.

To see that a Suslin tree $X$ is a counterexample, recall that a Suslin tree is a tree of height $\omega_1$ where every linearly ordered set is countable and every antichain is countable. Suppose we have a collection $(X_i)_{i \in I}$ of incomparable nonempty linearly ordered subsets of $X$. Let $x_i$ be the minimal element of $X_i$. This set $\{x_i \mid i \in I\}$ is an antichain, which must therefore be countable. Since each $X_i$ is countable, the supremum of the heights of the $X_i$ must be countable as well since $\omega_1$ is a regular cardinal.

Suslin trees generalize to higher levels and provide similar counterexamples higher up. It's harder to get rid of higher Suslin trees than it is to get rid of those with height $\omega_1$ but these are still conditional counterexamples since we can't prove the existence of such trees in ZFC.

For Question 2, if $n \geq 2$, it is necessary that $\xi$ is either $\omega$ or a weakly compact cardinal otherwise the result is false when $X$ is linearly ordered. The result is true for $\xi=\omega$ because of Ramsey's Theorem. There are two cases:

First, the easy case, if $X$ has a branch of height $\omega$ then apply the infinite form of Ramsey's Theorem to that branch.

Otherwise, by Kőnig's Lemma, the tree must have an infinite level. For simplicity, let's chop the base of the tree so that we can assume that the tree (or rather forest) has infinitely many roots. This gives a partition of $X$ into infinitely many rooted trees $(X_i)_{i \in I}$. Since $X$ has height $\omega$, the heights of these trees are unbounded and they are all finite. From a selection of these trees pick branches $Y_1,Y_2,\ldots$ so that the finite form of Ramsey's Theorem ensures that if $[Y_m]^n$ is $k$-colored then there is a homogeneous set $Z_m$ of size $m$ (where $n$ and $k$ are given by your setup). Apply Ramsey's Theorem to each $Y_m$ using your coloring and assemble the subchains $Z_m$ into a homogeneous tree of height $\omega$.

The same trick applies when $\xi$ is weakly compact. The basic idea is that a weakly compact cardinal satisfies the analogue of the infinite form of Ramsey's Theorem and it also has an analogue of Kőnig's Lemma. As a replacement for the finite form of Ramsey's Theorem, one needs to use the Erdős-Rado Theorem along with the observation that weakly compact cardinals are inaccessible.

Question 1 is true for countable $\omega^\xi$ but it is at least consistent that this is false at $\omega_1.$ There might be a ZFC counterexample but the first thing that popped to mind is a Suslin tree. It is consistent with ZFC that there is a Suslin tree but it is also consistent that there aren't any.

To see that a Suslin tree $X$ is a counterexample, recall that a Suslin tree is a tree of height $\omega_1$ where every linearly ordered set is countable and every antichain is countable. Suppose we have a collection $(X_i)_{i \in I}$ of incomparable nonempty linearly ordered subsets of $X$. Let $x_i$ be the minimal element of $X_i$. This set $\{x_i \mid i \in I\}$ is an antichain, which must therefore be countable. Since each $X_i$ is countable, the supremum of the heights of the $X_i$ must be countable as well since $\omega_1$ is regular.

For Question 2, if $n \geq 2$, it is necessary that $\xi$ is either $\omega$ or a weakly compact cardinal otherwise the result is false when $X$ is linearly ordered. The result is true for $\xi=\omega$ because of Ramsey's Theorem. There are two cases:

First, the easy case, if $X$ has a branch of height $\omega$ then apply the infinite form of Ramsey's Theorem to that branch.

Otherwise, by König's Lemma, the tree must have an infinite level. For simplicity, let's chop the base of the tree so that we can assume that the tree (or rather forest) has infinitely many roots. This gives a partition of $X$ into infinitely many rooted trees $(X_i)_{i \in I}$. Since $X$ has height $\omega$, the heights of these trees are unbounded and they are all finite. From a selection of these trees pick branches $Y_1,Y_2,\ldots$ so that the finite form of Ramsey's Theorem ensures that if $[Y_m]^n$ is $k$-colored then there is a homogeneous set $Z_m$ of size $m$ (where $n$ and $k$ are given by your setup). Apply Ramsey's Theorem to each $Y_m$ using your coloring and assemble the subchains $Z_m$ into a homogeneous tree of height $\omega$.

Question 1 is true for countable $\omega^\xi$ but it is at least consistent that this is false at $\omega_1.$ There might be a ZFC counterexample but the first thing that popped to mind is a Suslin tree. It is consistent with ZFC that there is a Suslin tree but it is also consistent that there aren't any.

To see that a Suslin tree $X$ is a counterexample, recall that a Suslin tree is a tree of height $\omega_1$ where every linearly ordered set is countable and every antichain is countable. Suppose we have a collection $(X_i)_{i \in I}$ of incomparable nonempty linearly ordered subsets of $X$. Let $x_i$ be the minimal element of $X_i$. This set $\{x_i \mid i \in I\}$ is an antichain, which must therefore be countable. Since each $X_i$ is countable, the supremum of the heights of the $X_i$ must be countable as well since $\omega_1$ is regular.

For Question 2, if $n \geq 2$, it is necessary that $\xi$ is either $\omega$ or a weakly compact cardinal otherwise the result is false when $X$ is linearly ordered. The result is true for $\xi=\omega$ because of Ramsey's Theorem. There are two cases:

First, the easy case, if $X$ has a branch of height $\omega$ then apply the infinite form of Ramsey's Theorem to that branch.

Otherwise, by Kőnig's Lemma, the tree must have an infinite level. For simplicity, let's chop the base of the tree so that we can assume that the tree (or rather forest) has infinitely many roots. This gives a partition of $X$ into infinitely many rooted trees $(X_i)_{i \in I}$. Since $X$ has height $\omega$, the heights of these trees are unbounded and they are all finite. From a selection of these trees pick branches $Y_1,Y_2,\ldots$ so that the finite form of Ramsey's Theorem ensures that if $[Y_m]^n$ is $k$-colored then there is a homogeneous set $Z_m$ of size $m$ (where $n$ and $k$ are given by your setup). Apply Ramsey's Theorem to each $Y_m$ using your coloring and assemble the subchains $Z_m$ into a homogeneous tree of height $\omega$.

The same trick applies when $\xi$ is weakly compact. The basic idea is that a weakly compact cardinal satisfies the analogue of the infinite form of Ramsey's Theorem and it also has an analogue of Kőnig's Lemma. As a replacement for the finite form of Ramsey's Theorem, one needs to use the Erdős-Rado Theorem along with the observation that weakly compact cardinals are inaccessible.