Yes, if $G$ is connected. Let $f$ be a function in $k[X]$ which satisfies a polynomial equation over $k[X]^G$. Everything in the $G$-orbit of $f$ must satisfy that polynomial equation. Thus the $G$-orbit is finite, because it is contained in the set of roots of a polynomial, and is connected because it is the orbit of a connected group action, so it is a single point, so $f\in k[X]^G$.

Yes, if $G$ is connected. Let $f$ be a function in $k[X]$ which satisfies a polynomial equation over $k[X]^G$. Everything in the $G$-orbit of $f$ must satisfy that polynomial equation. Thus the $G$-orbit is finite, because it is contained in the set of roots of a polynomial, and is connected because it is the orbit of a connected group action, so it is a single point, so $f\in k[X]^G$.

In fact, for $G$ not connected with $G_0$ the connected component of the identity and $G/G_0$ finite, the integral closure of $k[X]^G$ in $k[X]$ is $k[X]^{G_0}$ because any element in $k[X]^{G_0}$ has a finite orbit and the coefficients of the monic polynomial that vanishes exactly on the orbit are $G$-invariant.

Let $f$ be a function in $k[X]$ which satisfies a polynomial equation over $k[X]^G$. Everything in the $G$-orbit of $f$ must satisfy that polynomial equation. Thus the $G$-orbit is finite, because it is contained in the set of roots of a polynomial, and is connected because it is the orbit of a connected group action, so it is a single point, so $f\in k[X]^G$.

Yes, if $G$ is connected. Let $f$ be a function in $k[X]$ which satisfies a polynomial equation over $k[X]^G$. Everything in the $G$-orbit of $f$ must satisfy that polynomial equation. Thus the $G$-orbit is finite, because it is contained in the set of roots of a polynomial, and is connected because it is the orbit of a connected group action, so it is a single point, so $f\in k[X]^G$.