I am trying to solve the character table and some related questions. The questions are below, and what I have done is below that. Any help on any pieces I am sure will enlightening.

For parts c and d, I could really use some hints as I don't know where to begin.
e. The best I have figured out for the character table is
There are 5 conjugacy classes hence 5 irreducible representations. There 20 elements in the group and the only way to break this down to the sum of squares is $1^2 + 1^2 + 1^2 + 1^2 + 4^2$ so these are the orders of the conjugacy classes of the trivial element.
\begin{array}{rrrrrrrrrrr}
& C_1 & C_2 & C_3 & C_4 & C_5\\
\chi_0 & 1 & 1 & 1 & 1 & 1\\
\chi_1 & 1& 1& i & -i & -1 & \\
\chi_2 & 1 & 1 & -i & -i & -1& \\
\chi_2 & 1 &1 &-1 & -1& 1 & \\
\chi_3 & 4 & -1 & 0& 0& 0& \\
C_G(x) & 1 & 5 & 4 & 4 & 4 &
\end{array}
although I think $d,e$ should be a -i and i pair. With my calculations I am having a hard time just putting in numbers trying to satisfy $\left \langle \alpha, \beta\right \rangle := \frac{1}{|G|}\sum_{g \in G} \alpha(g) \overline{\beta(g)} $ and the column version of Schur's orthogonality relationship>
f. I see that the centralizer of $C_3$ has 4 elements but does this, and if so how, does this mean that elements $x \in C_3$ have order divisible by 4?
g. more properties of conjugacy classes and that somehow I can read this from the character table?
Really thanks for any hints, tips, or answers.