Is there a compact space with no countably generated dense subspace?

Ramiro, regarding 2), the set of all points in $I^\kappa$ with at most countably many non-zero coordinates is even Frechet-Urysohn, for every $\kappa$. This is due to Noble (see also exercise 3.10.D in Engelking).

Note: Frechet-Urysohn means that given a non-closed set $A$ and a point $x \in \overline{A} \setminus A $, there is a sequence inside $A$ converging to $x$.

An $\eta_1$-set is a linearly ordered set $(Q, \leq)$ with the property that if A and B are countable subsets of $Q$ satisfying $a < b$ for all $a \in A, b \in B$, then there is an $x \in Q$ satisfying $a < x < b$ for all $a \in A, b \in B$. (This definition and the notation are from Gillman and Jerison.) Let $(X, \leq)$ be the Dedekind compactification of an $\eta_1$-set, that is, $X$ is constructed from the $\eta_1$-set $Q$ in analogy to the construction of the reals from the rationals using Dedekind cuts, and then a first and last element is added. Then $X$ is a connected compact space. I can't recall a reference, but $X$ is the disjoint union of three dense sets: the set of P-points (which includes the endpoints), the set of points that are limits of strictly increasing sequences, and the set of points that are limits of strictly decreasing sequences. Furthermore, in $X$ a point is a limit of a non-trivial sequence if and only if it is an accumulation point of a countable set. If $D$ is a dense subset containing a P-point, it obviously does not have countable tightness. If $D$ is a dense subset containing a limit $x$ of a strictly increasing sequence, then $x$ is not a limit of a strictly decreasing sequence, so the set $S$ of elements of $D$ larger than $x$ contains $x$ in its closure, but no countable subset of $S$ has $x$ in its closure. Similarly, a dense subset containing a limit of a strictly decreasing set does not have countable tightness.