The probability that a screw is defective (screwed :P) is $0.01$, independently. Thus, probability of a screw not being defective is $(1-0.01) = 0.99$
The company offers a guarantee that in any pack of $10$, at most one is defective. That means, if there are more than one defective screws in your pack, the company will replace it.
The probability that there is more than one defective piece in a pack of $10$ screws can be calculated as:
$$\large \begin{align} P\left (\substack{\text{more than}\\\text{one defective}} \right ) \;\;\;&= \quad 1 - \Biggl (\; P \left ( \substack{\text{zero}\\\text{defective}} \right ) + \;P \left (\substack{\text{one}\\\text{defective}}\right )\; \Biggr )\\[2em] &=\quad 1 - \Biggl (\;\left (0.99^{10} \right ) \; + \; \underbrace{\left ( 10 \times 0.01 \times 0.99^9 \right )}_{\substack{\text{choose 1 screw out of 10}\\[0.5em]0.01 = P(1 \text{ defective screw})\\[0.5em]0.99^9 = P(9 \text{ fine screws})}}\; \Biggr )\\[2em] &\approx \quad 1-0.99573\\[1em] P\left (\substack{\text{more than}\\\text{one defective}} \right ) &\approx \quad 0.00427 \end{align}$$