Answer
a. $ \frac{p}{q}:\qquad \pm 1, \pm 5$
b. $1$ is a zero
c. $x\displaystyle \in\left\{\frac{3 \pm \sqrt {11}i}{2}, 1\right\}$
Work Step by Step
See The Rational Zero Theorem:
If $\displaystyle \frac{p}{q}$ is a zero of the polynomial $f(x) $with integer coefficients,
then $p$ is a factor of the constant term, $a_{0}$, and
$q$ is a factor of the leading coefficient, $a_{n}$.
------------------------
$f(x)=x^{3}-4x^{2}+8x-5$
a. Candidates for zeros, $\frac{p}{q}:$
$p:\qquad \pm 1, \pm 5$
$q:\qquad \pm 1$
$\displaystyle \frac{p}{q}:\qquad \pm 1, \pm 5$
b. Try for $x=1:$
$\begin{array}{lllll}
\underline{1}| & 1 & -4 & 8 & -5\\
& & 1 & -3 & 5\\
& -- & -- & -- & --\\
& 1 & -3 & 5 & |\underline{0}
\end{array}$
${1}$ is a zero,
$f(x)=(x-1)(x^{2} -3x+5)$
c. Using quadratic formula, $x=\frac{-b \pm \sqrt {b^2-4ac}}{2a}$.
we can solve for the quadratic $ax^2+bx+c$,
In our case of, $x^{2} -3x+5$, $a=1, b=-3, c=5$
$x=\frac{3 \pm \sqrt {-3^2-4\times 1 \times 5}}{2\times 1}$,
$x=\frac{3 \pm \sqrt {9-20}}{2}$,
$x=\frac{3 \pm \sqrt {11}i}{2}$
The zeros of f satisfy $f(x)=0$
$x\in\left\{\frac{3 \pm \sqrt {11}i}{2}, 1\right\}$
