I am currently studying for my algebra qualifying exams and always get tripped up by character tables (though it seems they should be much more straightforward). Here is a practice problem that I am fully stuck on.
My thought process is that
(a) we need to show that there exists nontrivial normal subgroups in the form of kernels of the characters presented. However it appears that $\chi_i(g) = \chi_i(1)$ is only true for $g=1$ for all the characters (ie. no nontrivial kernels)?
(b) one of the characters must correspond to a representation which can map into $GL_3(\mathbb{C})$ with kernel being 1.
Any help / reference materials to best learn the process of reading a table would be greatly appreciated! 
a) I'm not sure why you want to show there is a non-trivial normal subgroup? (by part (b) this actually isn't true). If $G$ is solvable then $[G,G]$ is a proper subgroup of $G$ so $G$ has non-trivial linear characters.
b) Two characters correspond to representations $\rho$ into $GL_3(\mathbb{C})$. Which ones? (Recall the dimension of the representation is $\chi(1)$). If an element $g$ is in the kernel of $\rho$, what is $\chi(g)$? This should show you that each of these $\rho$ is an injection.
c) Sylow's Thm says the number of subgroups of order $3$ (note the order of this group is $168=8\cdot 3\cdot 7$) divides $56$. Given that elements of order $3$ form a conjugacy class (and $g,g^{-1}$ form the same group), this leaves just one conjugacy class that could be the elements of order $3$ and therefore the number of subgroups of order $3$.