= Solution
A point $x$ of a set $S$ is an <extreme point> if $x=(1-t)y+tz$, with $y,z\in S$ and $0<t<1$, forces $y=z=x$. For a <convex set> it is enough to test midpoint decompositions: a nontrivial segment containing $x$ in its interior contains a smaller segment with midpoint $x$. The nonempty convex set $\mathbb R^2$ has no <extreme points>, since every $x$ is the midpoint of $x+e_1$ and $x-e_1$.
For the real <l-infinity sequence space>, a vector in the closed <unit ball> is extreme precisely when every coordinate is $1$ or $-1$. If $|x_j|<1$ for some $j$, choose $0<\delta\le1-|x_j|$. The distinct vectors $x\pm\delta e_j$ remain in the unit ball and have midpoint $x$, so $x$ is not extreme. Conversely suppose all coordinates have modulus one and $x=(y+z)/2$, with $\|y\|_\infty,\|z\|_\infty\le1$. In each real interval $[-1,1]$, an endpoint can be a midpoint only if both entries equal that endpoint. Thus $y_j=z_j=x_j$ for every $j$ and $y=z=x$. Therefore
$$
\boxed{\operatorname{ext}B_{\ell^\infty}=\{x:x_j\in\{-1,1\}\text{ for every }j\}.}
$$
For the real <absolutely summable sequence space>, any point with $\|x\|_1<1$ is nonextreme: choose a nonzero coordinate perturbation of norm less than $1-\|x\|_1$. If $\|x\|_1=1$ and two coordinates $x_j,x_k$ are nonzero, take $0<\delta<\min(|x_j|,|x_k|)$ and put
$$
h=\delta\bigl(\operatorname{sgn}(x_j)e_j-\operatorname{sgn}(x_k)e_k\bigr).
$$
Both perturbations preserve the signs of those two coordinates. One adds $\delta$ to one absolute value and subtracts it from the other, so $\|x+h\|_1=\|x-h\|_1=1$. Hence such an $x$ is not extreme. The only remaining candidates are $\pm e_j$. If $e_j=(y+z)/2$ with $\|y\|_1,\|z\|_1\le1$, its $j$th coordinate forces $y_j=z_j=1$. That consumes the entire norm allowance, forcing all other coordinates to vanish. The argument for $-e_j$ is identical. Thus
$$
\boxed{\operatorname{ext}B_{\ell^1}=\{e_j,-e_j:j\ge1\}.}
$$
For the real <l2 sequence space>, interior points of the unit ball again permit sufficiently small opposite perturbations and are not extreme. If $\|x\|_2=1$ and $x=(y+z)/2$, write $y=x+h$, $z=x-h$. The <parallelogram law> gives
$$
\frac{\|y\|_2^2+\|z\|_2^2}{2}=\|x\|_2^2+\|h\|_2^2=1+\|h\|_2^2.
$$
The left side is at most one, so $h=0$ and $y=z=x$. Consequently
$$
\boxed{\operatorname{ext}B_{\ell^2}=\{x:\|x\|_2=1\}.}
$$
These are the <extreme points of real sequence-space unit balls>: signs in $\ell^\infty$, coordinate spikes in $\ell^1$, and the whole sphere in $\ell^2$. The restriction to real scalars matters for the first description; the complex supremum ball has arbitrary unit-modulus coordinates instead.
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