Determine the number of elements of order 2 in AR 
So i have completed parts a and b. For b i reduced R to smith normal form and ended up with diagonals 1,2,6. From this i have said that the structure of the group is $Z_2 \oplus Z_6 \oplus Z$. But i have no idea what so ever about part c.

As stated in the comments, let us focus on $\,\Bbb Z_2\times\Bbb Z_6\,$ , which for simplicity (for me, at least) we'll better write multiplicatively as $\,C_2\times C_6=\langle a\rangle\times\langle b\rangle\,\;,\;\;a^2=b^6=1$ 
Suppose the first coordinate is $\,1\,$ , then the second one has to have order $\,2\,$ , and the only option is $\,(1,b^3)\,$ , so we go over elements with non-trivial first coordinate, and thus the second coordinate has to have order dividing two:
$$(a,1)\;,\;\;(a,b^3)$$
and that seems to be pretty much all there is: three involutions as any such one either has first coordinate trivial or not...
