= Solution
For $p=2$, the even residue class contains exactly one root, by the simple-root form of <Hensel lemma> at $a=0$: $f(0)=20$ and $f'(0)=-5$ is a unit. For the odd classes use the <strong form of Hensel lemma>:
$$
f(1)=16,\quad f'(1)=-2;\qquad f(3)=32,\quad f'(3)=22.
$$
In both cases $v_2(f(a))>2v_2(f'(a))=2$. Each gives exactly one root in $a+4\mathbb Z_2$. These two balls partition the odd integers, so there are no further roots.
For $p=3$, reduction gives $\overline f=X^3+X+2$, whose sole root is $2$. The derivative $3X^2-5$ is always a unit modulo $3$, so that root lifts uniquely. For $p=5$, the only possible residue is zero. But if $x\in5\mathbb Z_5$, then $x^3-5x\in25\mathbb Z_5$ while $20$ has valuation one, so $f(x)$ cannot vanish. \b[The exact root counts are]
$$
\boxed{\#\{f=0\text{ in }\mathbb Z_2\}=3,\quad \#\{f=0\text{ in }\mathbb Z_3\}=1,\quad \#\{f=0\text{ in }\mathbb Z_5\}=0.}
$$
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