Solutions

a)

• If $f$ and $g$ are both even, then $f(x)+g(x)=f(-x)+g(-x)$, so their sum is also even.
• Without loss of generality, let $f$ be even and $g$ be odd. Then $f(x)+g(x)\ne f(-x)-g(-x)$, so their sum is neither even nor odd.
• If $f$ and $g$ are both odd, then $f(x)+g(x)=-f(-x)-g(-x)$, so their sum is also odd.

b) If both $f$ and $g$ are odd, then

$$f(x)g(x)=(-f(x))(-g(x)),$$

so $fg$ is even. Likewise, if both $f$ and $g$ are even, then

$$f(x)g(x)=f(-x)g(-x),$$

so $fg$ is also even. Lastly, without loss of generality, if $f$ is even and $g$ is odd, then

$$f(x)g(x)=f(-x)(-g(-x))=-f(x)g(x).$$

So $fg$ is neither even nor odd, and this result applies more generally if one of $f$ and $g$ is even and the other function is odd.

c) If $g$ is even, then

$$f(g(x))=f(g(-x)).$$

Thus, $f\circ g$ is even whether $f$ is even or odd. If $g$ is odd and $f$ is even, then

$$f(g(x))=f(-g(-x)),$$

so the oddness of $g$ has no bearing on $f\circ g$, and their composition is even.

d) Let $f$ be any even function. Define the function $g$ as $g(x)=f(x)$ for all $x\ge 0$. Since $f$ is even, we have that for $x\ge 0$,

$$f(-x)=f(x)=g(x).$$

This means that the function $g$ we defined respects the equality $f(x)=g(|x|)$ for any $x\in\mathbb{R}$. Moreover, the equality holds if we were to have values defined over $(-\infty,0)$. Hence, we can construct infinitely many functions $h$ that satisfy $f(x)=h(|x|)$ by defining $h$ as

$$h= \begin{cases} f(x), & x\ge 0,\\ n, & x<0, \end{cases}$$

with infinitely many $n\in\mathbb{N}$.