A conjecture on 4-coloring maximal planar graphs
Conjecture 2 is already proved.
Quote from J.A. Tilley, The a-graph coloring problem(2017):

Theorem A.1. Let $G$ be an a-graph with boundary cycle $uxvy$ for the exterior 4-face and let $G$ have a 4-coloring $c$. Suppose, without loss of generality, that $c(x)=1$, $c(y)=1$ or 2, $c(u)=3$, and $c(v)=3$ or 4. Then there is either a 1–2 path between $x$ and $y$ or a 3–4 path between $u$ and $v$.


Proof. Suppose that G with 4-coloring c is a minimal counterexample to the theorem. Clearly G cannot have either an interior
xy edge or an interior u v edge. Let X be the set of vertices of G adjacent to x; they form an internal path between u and v that
includes at least one interior vertex of G. At least one of those interior vertices of G belonging to X is colored 2, for otherwise
the path between u and v would be colored 3–4, contradicting the supposition. Contract the various edges joining x to each
vertex of X colored 2, and change the color of x to 2. The result is a 4-colored a-graph F . The edge contractions do not create
any 3–4 path between u and v . By the minimality assumption, F must therefore have a 1–2 path between x and y. Reverse
the contractions and restore the color of x to 1 to reveal a 1–2 path between x and y in G, contradicting the supposition and
establishing the truth of the theorem.

Definiton: in the above paper, an a-graph is a plane graph such that one face has a 4-vertex cycle as boundary and all other faces have a triangle as boundary.