= Solution
The dual space $V^*$ is the <vector space> of <linear maps> $V\to\mathbb F$. If $B=(v_1,\ldots,v_n)$, its dual family satisfies $v_i^*(v_j)=\delta_{ij}$. Every functional $f$ obeys
$$
f=\sum_i f(v_i)v_i^*,
$$
so the family spans; evaluation on each $v_j$ proves linear independence. Thus it is a <basis> without any prior dimension argument.
For $p=a+bt+ct^2$,
$$
f_0(p)=a,\quad f_1(p)=a+b/2+c/3,\quad f_2(p)=a-b/2+c/3.
$$
The <basis> dual to $(f_0,f_1,f_2)$ is
$$
p_0=1-3t^2,\qquad p_1=t+\frac32t^2,\qquad p_2=-t+\frac32t^2,
$$
as direct substitution gives $f_i(p_j)=\delta_{ij}$. Hence the three functionals form a <basis> of $P_2^*$.
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