Your description of the weights of the standard rep is correct, you might be confused by the fact that $L_1+L_2+L_3=0$ on $\mathfrak{sl}_3$, so $L_1 = -L_2 - L_3$.
One way to approach this would be to write everything in terms of the two fundamental weights $\varpi_i$ (that is, the basis of $\mathfrak{h}^*$ dual to $h_1=\operatorname{diag}(1,-1,0)$ and $h_2=\operatorname{diag}(0,1,-1)$). What they are probably looking for is some statement of the form "if $V$ has highest weight $a\varpi_1 + b \varpi_2$ then $V^*$ has highest weight $f(a,b)\varpi_1 + g(a,b)\varpi_2$".
$v$ in your first displayed equation is called "a weight vector of $V$ with weight $\mu$". You might try proving a result that says that if you take a basis of $V$ consisting of weight vectors and $v$ is an element of this basis with weight $\mu$, then the element $v^*$ of the dual basis of $V^*$ has weight XXX....
Do you know how to draw the irreducible modules on the weight/root lattice? That might help you, since then you can see what the highest/lowest weights look like.
First a minor point: You're probably looking at representations of the complexification, because the $E_{ij}$ are not even in $\mathfrak{su}_n$; but they are in the complexified version $\mathfrak{su}_n \otimes \mathbb C$, which I will call $\mathfrak{g}$ from now on.
The crucial point is: When you write $E_{21}^2=0$, with the square "$^2$" you probably mean matrix multiplication. But that multiplication is not even part of the Lie algebra structure. Elements of the Lie algebra have no "squares" in that sense.
On the other hand, when your source writes "$(E_{21})^k v$", this is shorthand notation for $(\rho(E_{21}))^k (v)$, where $\rho: \mathfrak{g} \rightarrow End(V)$ is the actual representation on the vector space $V$; so now, $\rho(E_{21})$ is some endomorphism of $V$, and this endomorphism does have higher powers "$^k$" for $k \ge 2$, by which we mean $k$-fold composition of that endomorphism with itself.
So the problem is that we do not have $\rho(E_{21}^k)\stackrel{?}=\rho(E_{21})^k$, for the striking reason that the left hand side makes no sense to begin with. And even if it did, $\rho$ would have no reason to respect that kind of multiplication. The right hand side makes perfect sense though.
To see an example, look at the adjoint representation on $V=\mathfrak{g}$, where $\rho(E_{12}) = ad_{E_{12}}$, meaning that
$$\rho(E_{12}) (v) = [E_{12}, v] \quad \text{ hence}$$
$$\rho(E_{12})^2 (v) = [E_{12},[E_{12}, v]].$$
Can you find a $v$ in the Lie algebra $\mathfrak{g}$ such that this is $\neq 0$, showing that the endomorphism $\rho(E_{12})^2$ is non-zero?
Best Answer
Write $e_j$ for the column vector in $\mathbf{C}^n$ with a $1$ in position $j$ and zeros elsewhere. The action of a diagonal matrix $x$ on $e_j$ is then given by $$x \cdot e_j=L_j(x) e_j.$$ Now let $J=\{1 \leq j_1<j_2<\cdots<j_k \leq n\} \subseteq \{1,2,\dots,n\}$ be a subset of size $k$. The action of $x$ on the exterior product of the vectors $e_j$ for $j \in J$ is given by $$x \cdot e_{j_1} \wedge e_{j_2} \wedge \cdots \wedge e_{j_k}=\sum_{\ell=1}^k e_{j_1} \wedge \cdots \wedge x \cdot e_{j_\ell} \wedge \cdots \wedge e_{j_k}=\left(\sum_{\ell=1}^k L_\ell(x) \right) e_{j_1} \wedge e_{j_2} \wedge \cdots \wedge e_{j_k}.$$
This shows that the weights of the exterior product $\Lambda^k(\mathbf{C}^n)$ are precisely the sums $$L_J:=\sum_{j \in J} L_j$$ where $J$ ranges over subsets of $\{1,2,\dots,n\}$ of size $k$.