The first claim can be proved using Euler's formula: for a finite, connected planar graph with $v$ vertices, $e$ edges and $f$ faces, $v - e + f = 2$.
The second claim is the well-known four-color theorem. As noted by Johanna in the comments, it seems unlikely that you would be asked to prove this as an exercise.
The third claim can be proved by applying the fact that a tree with $n$ vertices has $n-1$ edges.
Ps. Alternatively, for the third claim, we can inductively prove the stronger claim that any tree with two or more vertices has at least two leaves.
Specifically, our induction assumption is that all trees with $v$ vertices, where $2 \le v \le n$, have at least two leaves. Clearly, this is vacuously true for $n=1$, so we can take that as our base case.
Now, consider a tree with $v = n+1$ vertices, and observe that deleting any edge of this tree leaves you with two smaller (but non-empty) trees. If one of these smaller trees has only one vertex, then it was a leaf in the original tree; otherwise, by the induction assumption, each of the smaller trees must have at least two leaves, and joining it to another tree with an edge can eliminate at most one of its leaves. Either way, the combined tree will inherit at least one leaf from each subtree, and thus has at least two leaves.
(Also note that this stronger claim is as strong as it can get: for any $n$, there exists an $n$-vertex tree with exactly two leaves. Exercise: what does such a tree look like?)