If the exponents are all odd, then $f(x)$ is the sum of odd functions, and hence is odd.
If the exponents are all even, then $f(x)$ is the sum of even functions, and hence is even.
As far as your last question, the sum of an odd function and even function is neither even nor odd.
Proof: Sum of Odd Functions is Odd:
Given two odd functions $f$ and $g$. Since they are odd functions $f(-x) = -f(x)$, and $g(-x) = -g(x)$. Hence: \begin{align*} f(-x) + g(-x) &= -f(x) - g(x) \\ &= -(f+g)(x) \\ \implies (f+g) & \text{ is odd if $f$ and $g$ are odd.} \end{align*}
Proof: Sum of Even Functions is Even:
Given two even functions $f$ and $g$, then $f(-x) = f(x)$, and $g(-x) = g(x)$. Hence: \begin{align*} f(-x) + g(-x) &= f(x) + g(x) \\ \implies (f+g) &\text{ is even if $f$ and $g$ are even.} \end{align*}
Proof: Sum of an odd function and even function is neither odd or even.
If $f$ is odd, and $g$ is even \begin{align*} f(-x) + g(-x) &= -f(x) + g(x) \\ &= -(f-g)(x) \\ \implies (f+g)&\text{ is neither odd or even if $f$ is even, $g$ is odd.} \end{align*}