I'm reading the Fulton and Harris Representation Theory book, trying to learn about Lie Algebras.
On pg. 213, they compute the killing form on $\mathfrak{h}^*$ for $\mathfrak{sl}_n(\mathbb{C})$. I understand the computation for $\mathfrak{h}$, and I know that for $h_1,h_2 \in \mathfrak{h}^*$, $B(h_1,h_2) = B(X,Y)$ where $h_1 = B(X,\cdot)$ and $h_2=B(Y,\cdot)$, but for the life of me I can't figure out how to compute the $\mathfrak{h}^*$ killing form?
Also, on pg. 162 when they discuss the representations of $\mathfrak{sl}_3(\mathbb{C})$, they say that since we know diagonalizable commuting matrices are simultaneously diagonalizable, using the Jordan Decomposition Theorem we know that $\rho(H)$ admits a direct sum decomposition of $V$ into eigenspaces. I know that by the Jordan Decomposition theorem, $H$ diagonalizable implies $\rho(H)$ is diagonalizable, but if $H_1$ and $H_2$ commute in $\mathfrak{h}$, how do I know $\rho(H_1)$ and $\rho(H_2)$ commute?
I don't have Fulton and Harris around right now so I'm not sure of exactly what you're looking at. However, for your first question, Killing forms are forms on Lie algebras, so there is a Killing form for $\mathfrak{h}$ (namely $\kappa(x,y) = \mathrm{tr}(\mathrm{ad}_x \circ \mathrm{ad}_y)$). By the "Killing form on $\mathfrak{h}^*$" the authors must be referring to the bilinear form induced by the Killing form on $\mathfrak{h}$ (which is induced in the way you indicated in your question: If $h_1=\kappa(x,\cdot)$ and $h_2=\kappa(y,\cdot)$, then $\kappa(h_1,h_2)=\kappa(x,y)=\mathrm{tr}(\mathrm{ad}_x \circ \mathrm{ad}_y)$). So to compute the "killing form on $\mathfrak{h}^*$" you merely need to compute the killing form on $\mathfrak{h}$ and use your isomorphism from $\mathfrak{h}$ to $\mathfrak{h}^*$: $h \mapsto \kappa(h,\cdot)$ to translate.
Your second question is a bit simpler to answer. Let $\rho:\mathfrak{g} \to \mathfrak{gl}(V)$ be a representation of some Lie algebra $\mathfrak{g}$ on some vector space $V$. Then for all $x,y \in \mathfrak{g}$, $\rho([x,y]) = [\rho(x),\rho(y)] = \rho(x) \circ \rho(y) - \rho(y) \circ \rho(x)$ (this is part of the definition of a representation). So if $[x,y]=0$ ($x$ and $y$ commute) you have $\rho(x) \circ \rho(y) - \rho(y) \circ \rho(x) = [\rho(x),\rho(y)] = \rho([x,y]) = \rho(0)=0$ so that $\rho(x) \circ \rho(y) = \rho(y) \circ \rho(x)$.
So since Cartan subalgebras like $\mathfrak{h}$ are Abelian Lie algebras, $\rho(\mathfrak{h})$ is a collection of commuting endomorphisms.