Are the adjoint representations of $\sla(n,\C)$, $\so(2n,\C)$, $\so(2n+1,\C)$, $\spa(n,\C)$ irreducible?
1. We use the root space $E\colon=\{x\in\R^n:\sum x_j=0\}$ with roots $e_j-e_k$, $j\neq k$ and base $b_1=e_1-e_2,\ldots,b_{n-1}=e_{n-1}-e_n$. Then for $j < k$ we get: $e_j-e_k=b_j+\cdots+b_{k-1}$ and $\l$ is dominant $\la\l,b_j\ra\geq0$, i.e. $\l_j-\l_{j+1}\geq0$, i.e. $j\mapsto\l_j$ is decreasing. Putting $\l\colon=b_1+\cdots+b_{n-1}=e_1-e_n$, we find that the adjoint representation of $\sla(n.\C)$ is a highest weight $\l$ representation.
2. The root space is $\R^n$ with roots $\pm(e_j+e_k)$ and $\pm(e_j-e_k)$, $j < k$. As base we choose for $j=1,\ldots,n-1$: $b_j=-e_j+e_{j+1}$ and $b_n\colon=e_1+e_2$. Then for $j < k$ we get: $-e_j+e_k=b_j+\cdots+b_{k-1}$. For $k > 2$:
$$
e_1+e_k=e_1+e_2-e_2+e_k=b_n+b_2+\cdots+b_{k-1}
$$
and for $j=2,\ldots,k-1$:
$$
e_j+e_k=(e_{j-1}+e_k)+b_{j-1}
$$
which gives by induction on $j=3,\ldots,k-1$:
$$
e_2+e_k=b_n+b_1+b_2+\cdots+b_{k-1},
e_j+e_k=b_n+b_1+2b_2+\cdots+2b_{j-1}+b_j+\cdots+b_{k-1}
$$
The highest weight:
$$
\l\colon=e_{n-1}+e_n=b_1+2(b_2+\cdots+b_{n-2})+b_{n-1}+b_n
$$
A weight $\l=\sum\l_je_j$ is dominant if $j\mapsto\l_j$ is increasing and $\l_1+\l_2\geq0$.
3. The root space is again $\R^n$ with roots $\pm(e_j+e_k)$ and $\pm(e_j-e_k)$, $j < k$ and $\pm e_j$, $j=1,\ldots,n$. For the base we choose $b_j=-e_j+e_{j+1}$ and $b_n\colon=e_1$. Then for $j < k$ we get: $e_j-e_k=b_j+\cdots+b_{k-1}$ and for $j < k$:
$$
e_j=b_j+e_{j+1}=b_1+b_2+\cdots+b_{j-1}+b_n,
e_j+e_k=2(b_1+\cdots+b_{j-1})+b_j+\cdots+b_{k-1}+b_n
$$
The highest weight:
$$
\l\colon=e_{n-1}+e_n=2(b_1+\cdots+b_{n-2})+b_{n-1}+b_n~.
$$
4. This is isometric to the dual root system of $\so(2n+1,\C)$ and thus a base is given by (cf. proposition: $b_j=-e_j+e_{j+1}$ and $b_n\colon=2e_1$. The highest weight is $2e_n$.