Yes; if $Y$ is any proper dense subspace of $X$, then there exists a vector $x \notin Y$. Set $Z$ to be the one-dimensional subspace spanned by $x$.
This can also happen with $Z$ infinite dimensional. Try $X = C([0,1])$, $Y$ the set of all polynomials, and $Z$ the set of functions which vanish on all of $[0,1/2]$. Then $Y$ is dense (by the Weierstrass approximation theorem), $Z$ is closed, and $Y \cap Z = \{0\}$ (since any nonzero polynomial can only vanish at finitely many points). This answers your updated question 2.
Edit: For your updated question 1, assuming the axiom of choice, there is a dense subspace whose intersection with any subspace of dimension at least two is nontrivial. Let $f : X \to \mathbb{R}$ be an unbounded linear functional (which one can construct given a Hamel basis) and let $Y$ be its kernel. $Y$ has codimension 1 so any two-dimensional subspace intersects it nontrivially. (More explicitly, if $x,y$ are linearly independent, then $f(x) y - f(y) x \in Y$ so the span of $x,y$ intersects $Y$.)
I claim $Y$ must also be dense. First, $Y$ is not closed; if it were, then $f$ would be continuous (standard fact). So $Y \subsetneq \bar{Y}$. But since $Y$ has codimension 1 the only subspace which properly contains it is $X$ itself. (To put it another way, let $y \in \bar{Y} \setminus Y$, so $f(y) \ne 0$. Given any $x \in X$, we write $x = (x - \frac{f(x)}{f(y)} y) + \frac{f(x)}{f(y)}y$ and we have shown that $x$ is in the span of $Y$ and $y$, hence is in $\bar{Y}$.)