Find the average value of $f$ on the given interval $f(x) = (x-3)^2$, $[2,5]$ After finding the average value of $f$, I need to find $c$ such that $f_\text{avg}= f(c)$
$$f(x) = (x-3)^2 \textrm{, }[2,5]$$
$$\begin{align}
f_\text{avg} &= \frac{1}{5-2} \int_{2}^{5} (x-3)^2dx \\
u = x-3, du = dx & ~~~~~ \frac{1}{3} \int_{2}^{5} u^2du = \frac{1}{3} \bigg[ \frac{1}{3}(x-3)^3 \bigg]_{2}^{5} \\
&= \frac{1}{3} \bigg(\frac{1}{3}(5-3)^3 - (\frac{1}{3}(2-3)^3\bigg) \\
&= \frac{1}{3} \bigg( \frac{8}{3} + \frac{1}{3} \bigg)\\
&= 3
\end{align}$$
The textbook says that $f_{avg} = 1$, not 3.
Did I mess up my u-substitution? 

Just the comments have pointed out, you miscalculated the last line. It should be 
$$\frac{1}{3}\cdot\Big(\frac{8}{3}+\frac{1}{3}\Big)=\frac{1}{3}\cdot\frac{9}{3}=1\neq3$$
To solve the second part, it should be an easy substitution
$$(x-3)^2=1 \iff x-3=\pm1 \iff x=4,2$$
