= Solution
The <dual vector space> is $V^*=\operatorname{Hom}_{\mathbb R}(V,\mathbb R)$, the <vector space> of all <linear functionals> on $V$. For a <vector subspace> $U\leq V$, its <annihilator of a vector subspace> is
$$
U^0=\{\phi\in V^*: \phi(u)=0\text{ for every }u\in U\}.
$$
Given a <basis> $x_1,\ldots,x_n$, define $x_i^*$ by $x_i^*(x_j)=\delta_{ij}$ and extend linearly. If $\sum_i a_i x_i^*=0$, evaluation at $x_j$ gives $a_j=0$, so the $x_i^*$ are <linearly independent>. Every $\phi\in V^*$ satisfies
$$
\phi=\sum_{i=1}^n\phi(x_i)x_i^*,
$$
so they also <span> $V^*$ and therefore form its <dual basis>.
Write $\phi=a x_1^*+b x_2^*+c x_3^*+d x_4^*$. The condition $\phi\in U^0$ is
$$
a+2b+3c+4d=0,\qquad 5a+6b+7c+8d=0.
$$
<Gaussian elimination> gives $b=-2c-3d$ and $a=c+2d$. Hence
$$
\boxed{U^0=\operatorname{span}\{x_1^*-2x_2^*+x_3^*,\;2x_1^*-3x_2^*+x_4^*\}.}
$$
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