I'm working out of Mumford's Red Book. In this question, a variety $X$ is complete if for all varieties $Y$, the projection morphism $p_2:X\times Y\to Y$ is closed.
I have two questions as following.
(1) if $X$ is complete, every morphism of varieties $f: \mathbb{A}_k^1 \rightarrow X$ can be extended to a unique morphism $\mathbb{P}_k^1 \rightarrow X$.
(2) Conversely, assume that every morphism $\mathbb{A}_k^1 \rightarrow X$ extends in a unique way to a morphism $\mathbb{P}_k^1 \rightarrow X$, can we conclude that $X$ is complete?
I have some idea about the first question. we can replace $\mathbb{P}_k^1$ by affine open cover $U_{x_i}$,$i=0,1$ and $U_{x_i}$ isomorphic the affine space, so we get a extension of $f$. But I do not know how to use the property of complete variety to prove this extend is unique.
We'll assume your definition of a complete variety is that the structure morphism $X\to\operatorname{Spec} k$ is proper. This is satisfied because proper = separated + universally closed + finite type: in your setting, a variety is separated and of finite type, and the extra assumption you add is universally closed.
Yes. This follows from the valuative criteria for properness applied to the discrete valuation ring $k[\frac{1}{t}]_{(\frac{1}{t})}\subset k(t)=k(\Bbb A^1_k)$.
No. Consider the affine part $X_0$ of an elliptic curve $X$ over $k$: any map $\Bbb A^1\to X_0$ can be upgraded to a map $\Bbb P^1\to X$ by the above, and then Riemann-Hurwitz tells us that this map must be constant. So the original map $\Bbb A^1\to X_0$ was constant, and this extends to a map $\Bbb P^1\to X_0$, but $X_0$ is not proper (proper + affine = finite, $X_0$ is one-dimensional).