Answer
See answers.
Work Step by Step
a. Check the sign of the Q-value, using data from Appendix B. We use atomic masses.
$$Q=\left( m_{\alpha}+m(^{14}_{7}Ni)-m(^{17}_{8}O)-m_{p} \right)c^2$$
$$ =\left(4.002603u+14.003074u-16.999132u-1.007825u\right)c^2\left( \frac{931.49MeV/c^2}{u}\right)$$
$$=-1.192MeV$$
This is negative, so it cannot occur unless there is some energy input. We are told that the incident KE is 9.85 MeV, which is enough. Yes, this reaction can occur.
b. The total KE of the products is the Q-value plus the KE brought in by the incoming alpha particle: -1.192MeV + 9.85 MeV = 8.66 MeV.