Note that odd/even-ness of a function must apply for all $x$ in the domain where the function is defined. Specific values don't necessarily mean anything in the grander context - to show a function is odd or even, it must be shown for arbitrary $x$.
Let $f(x) = x \cdot \cos(x)$. Consider $f(-x)$. Recalling cosine is an even function,
f(-x) &=-x \cdot \cos(-x) \\
&= -x \cdot \cos(x) \\
&= -f(x) \\
Thus, $f$ is odd, as $f(-x) = -f(x)$.
Let $g(x) = \tan^5 (x)$. Recall tangent is odd, and thus
g(-x) &= \tan^5 (-x)\\
&= (-1)^5 \tan^5 (x) \\
&= - \tan^5 (x) \\
&= - g(x)
(If instead you meant $\tan(x^5)$, the argument is pretty similar.)
Thus, $g$ is also odd.
It might also be worth noting:
You cannot say a function is "equal to" infinity. If $f(x) = 1/x$, $f(0)$ is not equal to $\infty$. In some contexts, it approaches infinity (as the right-hand limit as $x -> 0$), but $1/0 \neq \infty$.
This might be important to note when discussing even/odd-ness of such functions - the premise is that the function is defined in the first place. $\tan(x)$ is not defined for $x = k\pi$ for integers $k$. Not equal to infinity, or to negative infinity as might also be the case from a different viewpoint - just simply undefined.