Does there exist an even polynomial that has a term with an odd power?

Hint: You need $f(x) - f(-x)$ to be the zero polynomial.

The answer depends on which field you are working in. For example, over a field of characteristic $2$, every polynomial is even. Also, over a finite field $F_q$, we have that the polynomial $x^q - x$ is even (it vanishes for all $x\in F_q$) but its $x$ term has a non-zero coefficient.

However, over an infinite field of characteristic $\neq 2$, every even polynomial is a polynomial in $x^2$. This follows since $f(x) - f(-x) = 2g(x)$, where $g(x)$ consists of the odd powers of $x$ in $f$ and by assumption $2g(x)$ vanishes for all, and hence infinitely many, elements $x$. Thus $2g(x) = 0$ which implies $g(x) = 0$ (char $\neq 2$).

No. If $p(x)=a_nx^n+\cdots +a_1x+a_0$ then $$p(-x)=p(x)$$ $$a_0-a_1x+\cdots +(-1)^na_nx^n=a_0+a_1x+\cdots a_nx^n$$ Polynomials are identically equal exactly when their coefficients are equal, so $$a_1=-a_1\implies a_1=0,...$$ all coefficients of odd powers are $0$.

Alternatively, use induction. If $p$ is even, then $$-p'(-x)=p'(x)$$ so $p'$ is odd, and vice versa. Prove that constant and linear polynomials which are even and odd have only even and odd (respectively) terms, and you're done.

Alternate answer. The sum of two even functions is even. So if $p$ is an even polynomial, add scalar multiples of even powers of $x$ to eliminate the even powers in $p(x)$. You'll be left with a supposedly even polynomial that has only odd powers of $x$. But that makes what you have an odd function. So you have a function that is even and odd, and it's easy to prove such a function is the zero function. Therefore there were no odd terms left over after cancelling the even terms.