In how many ways can an inspector visit $4$ normal sites and $1$ "suspicious" one?

Your answer looks OK. Simplifying a little: $$\dfrac {10!}{2! \cdot 2! \cdot 2! \cdot 2!\cdot 2!}-\dfrac {9!}{2! \cdot 2! \cdot 2! \cdot 2! \cdot 1!} = \left(\frac{10}{2}-1\right)\frac{9!}{2^{4}} = 4\cdot\frac{9!}{16} = \frac{9!}{4}$$

Using a different approach, consider placing the low priority sites in order: $$\frac{8!}{2! \cdot 2! \cdot 2! \cdot 2!}$$

Then the bad site visits slot into the $9$ "gaps" in ${9\choose 2}$ ways: $$\frac{8!}{2! \cdot 2! \cdot 2! \cdot 2!}{9\choose 2} = \frac{8!}{16} \cdot\frac{9\cdot8}{2} =\frac{9!}{4} $$

The answer in the OP, $\dfrac {10!}{2! \cdot 2! \cdot 2! \cdot 2!\cdot 2!}-\dfrac {9!}{2! \cdot 2! \cdot 2! \cdot 2! \cdot 1!}$, correctly evaluates to $90, 720$. I had repeatedly made a computation mistake when evaluating this, which led me to believe it was wrong.