Answer
$(a) \quad$ f(x) can have 0,1, or 2 critical points. See examples below.
$(b)\quad$ f(x) either has two or no extreme values.
Work Step by Step
$f(x)=ax^{3}+bx^{2}+cx+d$
$f'(x)=3ax^{2}+2bx+c$ is a polynomial, so no critical points are caused by $f'(x)$ not being defined.
$f'(x)$ is a quadratic polynomial, and can have 0, 1, or 2 real zeros. This means that f can have 0,1, or 2 critical points.
Examples (see graphs below):
$f'(x)=x^{2}-1 \quad$ has two zeros.
A function with this derivative could be
$f(x)=\displaystyle \frac{x^{3}}{3}-x$
$f'(x)=(x-1)^{2}=x^{2}-2x+1 \quad\quad$ has one zero.
A function with this derivative could be
$f(x)=\displaystyle \frac{x^{3}}{3}-x^{2}+x$
Note that there are no extreme values in this case.
$f'(x)=x^{2}+1 \quad$ has no zeros.
A function with this derivative could be
$f(x)=\displaystyle \frac{x^{3}}{3}+x$
Note that there are no extreme values in this case.
$(b)$
From the above examples, we see that f(x) can have 2 or no extreme values. If it doesn't have two critical points, it has no extrema.