Does this graph contain $K_5$ or $K_{3,3}$ as subdivision or minor? Does this graph contain subdivision of $K_5$ or $K_{3,3}$?
Does this graph contain  $K_5$ or $K_{3,3}$ as minor?

I'm not sure if I'm correct, but I think the answer is yes for both questions. $K_5$ has 5 vertices of degree 4, this graphs has 6 of them, and $K_{3,3}$ has 6 vertices of degree 3, and this graph also has 5 of them. are these enough to conclude that this graph says $G$ contain both $K_5$ and $K_{3,3}$ as subdivision and minor?
 A: Denote the six inner vertices as $A,B,C,D,E,F$ in anticlockwise order. Denote the outer vertices as $A',B',C',D',E',F'$ accordingly.
Let's try subdivision of $K_5$: To find such, we need five vertices of degree (at least) $4$. Since there are six such vertices, we have littel choice. In fact, by symmetry, all possible choices are the same, so we may pick $A,B,C,D,E$. Then to exhibit a subdivision, we need to have a path from each of these vertices to each other vertex - and these paths must be vertex-disjoint (apart from the end points)!
Since the vertices are only of degree $4$, we must use all edges originating from our chosen vertices. This already completes $6$ of our needed $10$ paths: $AC, AD, AE, BD, BE, CE$. 
Of the paths from $C$ to $B$ and from $C$ to $D$, one must make use of $C'$ and one must ake use of $F$.
Then for the path from $A$ to $B$ we must take  $AA'B'B$ because $F$ is already used.  Similarly we have $EE'D'D$. But then the path from $C$ to $B$ or $D$ that goes via $C$ cannot be comppleted.  We conclude that this graph does not contain a subdivision of $K_5$ as subgraph.
A: Label the outer vertices clockwise $A,B,C,D,E,F$, the inner vertices $A',B',C',D',E',F'$
(starting at the same ray from the center).
Now there is a subdivision of $K_{3,3}$ having one partite set $A,C',E$ and the other $A',C,F$.
The connections are
$AA'$, $ABC$, $AF$
$C'A'$, $C'C$, $C'F'F$
$EE'A'$, $EDC$, $EF$
A: Counting the number of vertices of degree $4$ or $3$ is not sufficient.  Consider the below graph with two concentric circles and a zigzag between them.  Each vertex has degree $4$ and there are $12$ of them.  This graph is planar, so includes neither $K_6$, nor $K_{3,3}$  

A: 
Subgraphs...
I think I can see a $K_{3,3}$ straight off.... I strongly suspect from a casual viewing that there is no $K_5$. But your count of vertex degrees doesn't determine either matter.
If you can find 5 vertices of degree 4 that are linked, you have a $K_5$ - otherwise not. However in this case you can first (and iteratively) discard any vertices that  are degree 3 or below. The first pass of discarding low-degree vertices removes the outer ring, and the second pass leaves an empty graph - there is no $K_5$.
$K_{3,3}$ requires cycles of length 4, which the outer ring vertices do not have, so discard them....  Then inner six vertices do not form $K_{3,3}$, they form a triangular prism - I was mistaken earlier.

So much for my foray into subgraphs, which I think I mistakenly wandered into based on the vertex degree discussion. The question actually asked about subdivision or minor, for which counting vertex degree is even more irrelevant. 

Minors
As Henning correctly points out, the graph has a minor of $K_6$ - which can be formed by contracting the 6 edges between the outer ring and the inner figure - and so also contains minors of $K_5$ and $K_{3,3}$.

Subdivisions
The graph does contain subdivisions of $K_{3,3}$. These can can be produced by deleting two opposite nodes from the outer ring and smoothing out the other outer-ring nodes. There are also two inner-figure edges to remove to get to $K_{3,3}$. Hagen's answer demonstrates that there is no subdivision of $K_5$ in this graph.
A: the answer is yes. For more defails see following literature:
http://www.math.ucla.edu/~mwilliams/pdf/petersen.pdf
