Find for which $n\in \mathbb N$ exist a base of space $\{p_0,p_1,p_2,p_3,\cdots,p_n\}$ of space $P_n$ such that:
a) none of polynomials does not have degree two
b) every polynomial have even degree
c) every polynomial have odd degree
I do not uderstand what he want , for a) dimension is n, for b and c dimension is $\frac{(n+1)}{2}$
Answer to the original question:
$(a)$ Your solution is correct, but it is not defined for $y = 0$. You can add something like:
$$f(x,y) = \begin{cases} \frac{x^2}y, &\text{ if }y \ne 0 \\ 0, &\text{ if } y = 0\end{cases}$$
$(b)$ Consider complex conjugation $z \mapsto \overline{z}$.
Answer to the new question:
$(a)$ If $\deg p_i = 2, \forall i$ then $P_n = \operatorname{span}\{p_0, \ldots, p_n\} \subseteq P_2$ so the only candidate is $n = 2$. Indeed, $\{1+x^2, x+x^2, x^2\}$ is a basis for $P_2$.
$(b)$ If all $p_i$ are of even degree, then clearly $n$ cannot be odd because then $\deg p_i \le n-1, \forall i$ so $P_n = \operatorname{span}\{p_0, \ldots, p_n\} \subseteq P_{n-1}$. If $n$ is even then $$\{1+x^n, x+x^n, \ldots, x^{n-1} + x^n, x^n\}$$ is a basis for $P_n$ with $\deg p_i = n$ which is even.
$(c)$ Similarly we see that $n$ must be odd.