First, we shall prove two following lemmas:

Lemma 1. If $a$ is an integer number that is not a perfect cube, and $m, n, p$ are integer numbers such that $m+n \sqrt[3]{a}+p \sqrt[3]{a^{2}}=0$, then
$$m=n=p=0 .$$
Proof of Lemma 1. Since $m+n \sqrt[3]{a}+p \sqrt[3]{a^{2}}=0$, then $\sqrt[3]{a}$ is the root of $P(x)=p x^{2}+n x+m$ with integer coefficients.
Consider the polynomial $Q(x)=x^{3}-a$, then $\sqrt[3]{a}$ is also a root of $Q(x)$. Assume that $Q(x)$ is reducible, then one of the factors must have degree 1, that is $Q(x)$ has a rational root. But we know that $\sqrt[3]{a}$ is irrational since $a$ is not a perfect cube (by assumption). Hence, $Q(x)$ is irreducible.
In other words, $Q(x)$ is the minimal polynomial of $\sqrt[3]{a}$, and this leads to $P(x)=0$, or $m=n=p=0$.

Lemma 2. Given that $P(x)$ is a polynomial with integer coefficients, and $a$ is an integer such that $\sqrt[3]{a}$ is not an integer. If
$$P\left(\sqrt[3]{a^{2}}+\sqrt[3]{a}\right)=m+n \sqrt[3]{a}+p \sqrt[3]{a^{2}}$$
then $a-1 \mid n-p$.

Proof of Lemma 2. We will prove by induction on the degree of $P(x)$ which is denoted by $k$.
For $k=1$, it is trivial.
Assume that the statement is true for $k=t$, consider $P(x)$ of degree $t+1$, we write $P(x)=x Q(x)+s$, with $Q(x)$ of degree $t$ and $s=P(0)$ is an integer.
Assume that $Q\left(\sqrt[3]{a^{2}}+\sqrt[3]{a}\right)=d+e \sqrt[3]{a}+f \sqrt[3]{a^{2}}$, then by induction hypothesis $a-1 \mid e-f$. This gives
$$\begin{aligned} P\left(\sqrt[3]{a^{2}}+\sqrt[3]{a}\right) & =\left(\sqrt[3]{a^{2}}+\sqrt[3]{a}\right)\left(d+e \sqrt[3]{a}+f \sqrt[3]{a^{2}}\right)+s \\ & =s+e a+f a+(d+f a) \sqrt[3]{a}+(d+e) \sqrt[3]{a^{2}} \end{aligned}$$
Hence, $(d+f a)-(d+e)=f a-e=(f a-f)+(f-e)$, which is a multiple of $a-1$.

Now we turn to the problem.

For part a), we show that $P(x)$ does exist, for example we choose $P(x)= u x^{2}+v x+w$, where $u, v, w$ must satisfy
$$u+v=220,\;5 u+v=284,\;10 u+w=0 .$$
That is, $P(x)=16 x^{2}+204 x-160$.

For part b), since $1210-1184=26$, which is not a multiple of $4$, in this case there is no $P(x)$ satisfying the condition of the problem.