Here's an idea for the first question. I'm really thinking over $\mathbb C$. Maybe someone who knows enough can make it into a proof. I think that if $X$ has degree $D$ then $n$ can be taken to be $\frac{(D-1)(D-2)}{2}$, the genus of a smooth plane curve of degree $D$. The idea is to show that each $n$-path-component is open.

Let $X$ be $d$-dimensional. Let $a\in X$ be a point. Choose a linear projection $\pi$ to $\mathbb P^{d+1}$ such that it embeds a neighborhood $U$ of $a$, and such that $U=\pi^{-1}(\pi(U))$. By intersecting $\pi(X)$ with $2$-dimensional planes through $a$ we see that for any point $b\in X$ sufficiently close to $a$ there is an irreducible curve of genus at most $n$ mapping into $\pi(X)$ such that $\pi(a)$ and $\pi(b)$ are in its image. Lift so that the curve maps to $X$.

Edit: This is wrong. I forgot that we were seeking smooth curves in a smooth variety.

Edit: Let's start again. Taking a cue from Misha's answer: Suppose $V$ is quasiprojective, smooth, and connected, and of degree $d$. Let $p$ and $q$ be distinct points in $V$. If $dim(V)>1$, then there is a hyperplane $H$ through $p$ and $q$ that is transverse to $V$. (I'll discuss this claim below.) Now $V\cap H$ is smooth and of dimension $d-1$ (by transversality) and connected (by Lefschetz, since $dim(V)>1$) and again of degree $d$. Repeat until you have reduced the dimension to one. Now you have a smooth curve of degree $d$ in some projective space, so its genus is at most $\frac{(d-1)(d-2)}{2}$.

Now, why did that hyperplane exist? Let $L$ be the line determined by $p$ and $q$, and let $P$ be the projective space of all hyperplanes containing $L$. To make $H\in P$ transverse to $V$ at $p$, we just need to avoid a proper closed subspace of $P$, a projective space of codimension $dim (V)$, or $dim(V)-1$ if $L$ is tangent to $V$ at $p$. Likewise to make $H\in P$ transverse to $V$ at $p$ we just need to avoid another such subspace of $P$. Let $U\subset P$ be the remaining open subset. To make $H\in U$ transverse to $V-\lbrace p,q\rbrace$, consider the space of all pairs $(H,x)$ with $H\in U$ and $x\in H-\lbrace p,q\rbrace$. The projection $(H,x)\mapsto x$ to the complement of $\lbrace p,q\rbrace$ is a submersion, so the inverse image of $V-\lbrace p,q\rbrace$ is smooth. The projection $(H,x)\mapsto H$ of this smooth thing to $U$ is transverse to some point $H$. This $H$ is then transverse to $V$.