Answer
a) $2n$, where $n$ is the degree of $f$
b) $n^2$
c) $m+n$, where $m$ is the degree of $g$
d) $mn$
Work Step by Step
Let's note the functions:
$f(x)=a_nx^n+a_{n-1}x^{n-1}+....+a_1x+a_0$
$g(x)=b_mx^m+b_{m-1}x^{m-1}+....+b_1x+b_0$
a) $(f\cdot f)(x)=f(x)\cdot f(x)$
$=(a_nx^n+a_{n-1}x^{n-1}+....+a_1x+a_0)\cdot (a_nx^n+a_{n-1}x^{n-1}+....+a_1x+a_0)$
$=a_n^2x^{2n}+.....+a_0^2$
Therefore the degree of $f\cdot f$ is $2n$.
b) $(f\circ f)(x)=f(f(x))=f(a_nx^n+a_{n-1}x^{n-1}+....+a_1x+a_0)$
$=a_n(a_nx^n+a_{n-1}x^{n-1}+....+a_1x+a_0)^n+....+a_0$
$=a_n(a_n^nx^{n^2}+....)+....+a_0$
$=a_n^{n+1}x^{n^2}+.......+a_0$
Therefore the degree of $f\circ f$ is $n^2$.
c) $(f\cdot g)(x)=f(x)\cdot g(x)$
$=(a_nx^n+a_{n-1}x^{n-1}+....+a_1x+a_0)\cdot (b_mx^m+b_{m-1}x^{m-1}+....+b_1x+b_0)$
$=a_nb_mx^{n+m}+.....+a_0b_0$
Therefore the degree of $f\cdot g$ is $m+n$.
d) $(f\circ g)(x)=f(g(x))=f(b_mx^m+b_{m-1}x^{m-1}+....+b_1x+b_0)$
$=a_n(b_mx^m+b_{m-1}x^{m-1}+....+b_1x+b_0)^n+....+a_0$
$=a_nb_m^nx^{mn}+....+a_0$
Therefore the degree of $f\circ g$ is $mn$.
