How do I know if these degree sequences are planar or non planar? I'm trying to prove these two degree sequences are planar and non planar. How do I do that? I learned of $m \le 3n-6$, but that doesn't always work. Example is $K_{3,3}$; if you tested this graph with this formula, it will say that it is planar even though $K_{3,3}$ is not. 
The degree sequence of $G$ is $(4,4,4,5,5,5,6,6,6,7,7,7)$. Prove that G is non planar.
The next degree sequence is $(1,2,2,3,4,4,4)$. Prove that G is planar. 
How do I prove and disprove these two problems?
 A: If a graph $G$ is planar, then $|E(G)|\leq 3|V(G)|-6$, as a consequence of Euler's formula. Equivalently if $G$ does NOT satisfy $|E(G)|\leq 3|V(G)|-6$, then the graph is non-planar. However, the converse isn't true, as you noticed with $K_{3,3}$. Just because the inequality holds does not mean the graph is planar. 
For the first sequence, the graph has $12$ vertices and $(4+4+4+5+5+5+6+6+6+7+7+7)/2 = 33$ edges (by the Handshaking Lemma). Then you have $3v-6 = 30 < 33 = e$, so the graph doesn't satisfy $e\leq 3v-6$ and cannot be planar.
For the second sequence, you will have to use a different argument because that inequality cannot help you to prove a graph is planar. This may be breaking out the heavy machinery for this problem, but Kuratowski's Theorem should work. The graph has 7 vertices and $(1+2+2+3+4+4+4)/2 = 10$ edges. The vertex of degree 1 can be removed from the graph without affecting whether it is planar or not. So removing the vertex of degree 1, you're left with a graph with 6 vertices and 9 edges, but it also has two vertices of degree at most 2. Thus this graph is not $K_{3,3}$, nor can it have $K_{3,3}$ or $K_5$ as a minor (because it only has 9 edges). So the graph must be planar.
