I labeled the nodes to make my explanation easier.

You do indeed have a Petersen graph, though the inner part is rotated a bit from the examples online. The nice thing about this graph is that it's symmetrical. That means you can (at least in some cases) examine just one of the outer nodes and one of the inner nodes, so you don't have to try every single possibility. The graph is also small enough for you to figure these out by hand. I agree with the numbers, but I'll explain how I figured it out.

Girth I checked manually, with a sort of breadth-first approach, and took advantage of symmetry.

Distance from A (from node -> to node)

Code:

A->B 1
A->E 1
A->G 1
B->C 2
B->H 2
E->D 2
E->F 2
G->I 2
G->J 2
C->D 3
C->I 3
H->F 3
H->J 3
...

You could carry that on until you map all paths back to A, and work backwards though the list, picking the lowest cost routes to each node. I actually stopped early because I realized that, from node C, which was distance 2 from A, I could get to I, which was also distance 2 from A. There was just the cost of moving from C to I, which was cost 1. There was no way to get from a vertex that was distance 2 from A, to a vertex that was distance 1 from A. That gave me g(G) = dist(A->C) + dist(A->I) + dist(C->I) = 2+2+1 = 5. The fact that every node was reachable (not necessarily in a cycle) from A in at most 2 edges made me realize that in this graph, no node is more than 2 edges away from any other node. I didn't know if I would need that info yet, but it couldn't hurt. that is not necessary, since each cycle from A includes at least one of the inner nodes (F-J)

Next, onto clique number. It helped me immensely to look at the chart of complete graphs here. I could easily rule out the larger cliques (5 and above), since each node would need ω(G)-1 edges, and the nodes in this graph all have only 3 edges. That means the biggest clique possible might be 4. But we see that, for any clique size 3 or larger, there must be a triangle in the graph. This graph has no triangle. Thanks to symmetry, this is easy to check. Take node A, it has 3 neighbors, B, E and G. To make a triangle, two of those neighbors must share an edge. Similarly, examining an "inner" node, F, you find that none of its neighbors (E, H or I) share an edge. Thus, the best we can do is ω(G) = 2.

Lastly, is the chromatic number. I assume you're talking about vertex coloring since edge coloring is usually specified explicitly. There happens to be a fancy formula for calculating the number of t-colorings for graphs, where t is the number of different colors you use. Wikipedia shows the formula for Petersen graphs here. You can see that if t is 0, 1 or 2, one of the first terms will be 0, and there will be no colorings. Plugging 3 into the equation yields a positive result (120 different colorings I think it was), so χ(G) = 3. Intuitively, you can rule out χ(G) = 1, since there are nodes with neighbors. Also, you can rule out χ(G) = 2, by trial and errors, using alternating colors. So A is red, B, E and G are green. Then their neighbors (C, D, F, I and J) are red again. But C and D are themselves neighbors, and are both red, thus the graph is not 2-colorable. If you try 3 colors, you should be able to do it pretty easily.