Yes! (I assume it was implicit in your question that the variety be projective?)

More generally: any smooth, complex, rationally connected projective variety is simply connected. See Debarre's book ("Higher dimensional algebraic geometry").

Alternatively, take a look at Debarre's Bourbaki talk ("Varietes rationnellement connexes"). The idea is that a rationally connected variety has no holomorphic forms, so by Hodge theory the structure sheaf $\mathcal{O}_X$ is acyclic, implying $\chi(X,\mathcal{O}_X) = 1$. If $f : Y \to X$ is a connected etale cover, then $Y$ is again simply connected, so by the same argument $\chi(Y,\mathcal{O}_Y) = 1$, so $\deg{f} = 1$. So $X$ is simply connected.

In positive characteristic the Hodge theory fails, so the argument as such doesn't stand, but you may still get the simple connectedness as a consequence of the fibration theorem of Graber-Harris-Starr-de Jong. See the mentioned Bourbaki expose.

Yes! (I assume it was implicit in your question that the variety be projective?)

More generally: any smooth, complex, rationally connected projective variety is simply connected. See Debarre's book ("Higher dimensional algebraic geometry").

Alternatively, take a look at Debarre's Bourbaki talk ("Varietes rationnellement connexes"). The idea is that a rationally connected variety has no holomorphic forms, so by Hodge theory the structure sheaf $\mathcal{O}_X$ is acyclic, implying $\chi(X,\mathcal{O}_X) = 1$. If $f : Y \to X$ is a connected etale cover, then $Y$ is again rationally connected, so by the same argument $\chi(Y,\mathcal{O}_Y) = 1$, so $\deg{f} = 1$. So $X$ is simply connected.

In positive characteristic the Hodge theory fails, so the argument as such doesn't stand, but you may still get the simple connectedness as a consequence of the fibration theorem of Graber-Harris-Starr-de Jong. See the mentioned Bourbaki expose.

Yes! (I assume it was implicit in your question that the variety be projective?)

More generally: any smooth, complex, rationally connected projective variety is simply connected. See Debarre's book ("Higher dimensional algebraic geometry").

Yes! (I assume it was implicit in your question that the variety be projective?)

More generally: any smooth, complex, rationally connected projective variety is simply connected. See Debarre's book ("Higher dimensional algebraic geometry").

Alternatively, take a look at Debarre's Bourbaki talk ("Varietes rationnellement connexes"). The idea is that a rationally connected variety has no holomorphic forms, so by Hodge theory the structure sheaf $\mathcal{O}_X$ is acyclic, implying $\chi(X,\mathcal{O}_X) = 1$. If $f : Y \to X$ is a connected etale cover, then $Y$ is again simply connected, so by the same argument $\chi(Y,\mathcal{O}_Y) = 1$, so $\deg{f} = 1$. So $X$ is simply connected.

In positive characteristic the Hodge theory fails, so the argument as such doesn't stand, but you may still get the simple connectedness as a consequence of the fibration theorem of Graber-Harris-Starr-de Jong. See the mentioned Bourbaki expose.