Answer
$(a)$ Yes
$(b)$ Not necessarily
$(c)$ $h$ is odd
$(d)$ $h$ is even
Work Step by Step
$h=f◦g$
$(a)$ Let's consider the following:
$g(x)=g(-x)$
$f(g(-x))=f(g(x))$
$h(x)=h(-x)$
As we can see, if $g$ is even then the $h$ is also even function.
$(b)$ Let's consider the following:
$g(x)=-g(-x)$
$f(g(x))=f(-g(-x))$
$f(-g(-x))\ne -f(g(-x))$
$h(x)\ne-h(-x)$
As we see, if $g$ is odd, then $h$ is not necessarily odd.
$(c)$
$g(x)=-g(-x)$
$f(x)=-f(-x)$
$f(g(x))=f(-g(-x))$
$f(-g(-x))=-f(g(-x))$
$h(x)=-h(-x)$
As we can see, if $g$ and $f$ are both odd functions, then $h$ is also an odd function.
$(d)$
$g(-x)=-g(x)$
$f(x)=f(-x)$
$f(g(x))=f(g(-x))$
$f(g(-x))=f(-g(x))$, since g is odd
$h(x)=h(-x)$
As we can see, if $g$ is odd and $f$ is even, then $h$ is also an even function.
