Yes, rationally connected varieties in characteristic zero are uniruled (a fact which follows essentially by definition).

One can think about this also by exploiting the following characterization of uniruled and rationally connected varieties, that can be found in Debarre's book Higher dimensional algebraic geometry, Chapter 4.

Let $X$ be a smooth projective variety of dimension $d$, defined over a field of characteristic zero. Recall that a rational curve $f \colon \mathbb{P}^1 \to X$ is called $r$-free if $f^*T_X \otimes \mathscr{O}_{\mathbb{P}^1}(-r)$ is generated by its global sections. By Grothendieck' s Theorem any vector bundle over $\mathbb{P}^1$ splits as a direct sum of line bundles, so $r$-free means $$f^*T_X = \mathscr{O}_{\mathbb{P}^1}(a_1) \oplus \ldots \oplus \mathscr{O}_{\mathbb{P}^1}(a_d),$$ with $a_i \geq r$ for every $i$. Now one has the following

Proposition. $X$ is uniruled if and only through a general point $x \in X$ there is a $0$-free rational curve. Moreover $X$ is rationally connected if and only if through a general point $x \in X$ there is a $1$-free rational curve.

Yes, rationally connected varieties are uniruled (a fact which follows essentially by definition).

In characteristic zero, one can think about this also by exploiting the following characterization of uniruled and rationally connected varieties, that can be found in Debarre's book Higher dimensional algebraic geometry, Chapter 4.

Let $X$ be a smooth projective variety of dimension $d$, defined over a field $\mathbb{K}$ of characteristic zero. Recall that a rational curve $f \colon \mathbb{P}^1 \to X$ is called $r$-free if $f^*T_X \otimes \mathscr{O}_{\mathbb{P}^1}(-r)$ is generated by its global sections. By Grothendieck' s Theorem any vector bundle over $\mathbb{P}^1$ splits as a direct sum of line bundles, so $r$-free means $$f^*T_X = \mathscr{O}_{\mathbb{P}^1}(a_1) \oplus \ldots \oplus \mathscr{O}_{\mathbb{P}^1}(a_d),$$ with $a_i \geq r$ for every $i$. Now one has the following

Proposition. Assume $\textrm{char}(\mathbb{K})=0$. Then $X$ is uniruled if and only through a general point $x \in X$ there is a $0$-free rational curve. Moreover $X$ is rationally connected if and only if through a general point $x \in X$ there is a $1$-free rational curve.

Remark. The condition $\textrm{char}(\mathbb{K})=0$ is an essential one. For instance, if $\textrm{char}(\mathbb{K})=p>0$ then the Fermat hypersurface of degree $p^r+1$ in $\mathbb{P}^N$, with $N \geq 4$ and $r \geq 1$, is uniruled by lines, none of which are free. Moreover, if $p^r >N$ any such a hypersurface has ample canonical class, then it contains no free rational curves at all.

The answer is yes and it is a well known standard fact in Algebraic Geometry.

One can think about this topic also by looking at the following characterization of uniruled and rationally connected varieties, that can be found in Debarre's book Higher dimensional algebraic geometry, Chapter 4.

Let $X$ be a smooth projective variety of dimension $d$, defined over a field of characteristic zero. Recall that a rational curve $f \colon \mathbb{P}^1 \to X$ is called $r$-free if $f^*T_X \otimes \mathscr{O}_{\mathbb{P}^1}(-r)$ is generated by its global sections. By Grothendieck' s Theorem any vector bundle over $\mathbb{P}^1$ splits as a direct sum of line bundles, so $r$-free means $$f^*T_X = \mathscr{O}_{\mathbb{P}^1}(a_1) \oplus \ldots \oplus \mathscr{O}_{\mathbb{P}^1}(a_d),$$ with $a_i \geq r$ for every $i$. Now one has the following

Proposition. $X$ is uniruled if and only through any point $x \in X$ there is a $0$-free rational curve. Moreover $X$ is rationally connected if and only if through any point $x \in X$ there is a $1$-free rational curve.

Yes, rationally connected varieties in characteristic zero are uniruled (a fact which follows essentially by definition).

One can think about this also by exploiting the following characterization of uniruled and rationally connected varieties, that can be found in Debarre's book Higher dimensional algebraic geometry, Chapter 4.

Let $X$ be a smooth projective variety of dimension $d$, defined over a field of characteristic zero. Recall that a rational curve $f \colon \mathbb{P}^1 \to X$ is called $r$-free if $f^*T_X \otimes \mathscr{O}_{\mathbb{P}^1}(-r)$ is generated by its global sections. By Grothendieck' s Theorem any vector bundle over $\mathbb{P}^1$ splits as a direct sum of line bundles, so $r$-free means $$f^*T_X = \mathscr{O}_{\mathbb{P}^1}(a_1) \oplus \ldots \oplus \mathscr{O}_{\mathbb{P}^1}(a_d),$$ with $a_i \geq r$ for every $i$. Now one has the following

Proposition. $X$ is uniruled if and only through a general point $x \in X$ there is a $0$-free rational curve. Moreover $X$ is rationally connected if and only if through a general point $x \in X$ there is a $1$-free rational curve.