According to MO theory the bond orders for O22-, CO and NO+ respectively, are

🧠 All three species are 14-electron molecules — isoelectronic with $\ce{N2}$. Apply MO theory uniformly.

$\ce{O2^{2-}}$ (18e): \sigma_{1s}^2\sigma^*_{1s}^2\sigma_{2s}^2\sigma^*_{2s}^2\sigma_{2p}^2\pi_{2p}^4\pi^*_{2p}^4 BO = (10 − 8)/2 = 1

CO (14e): Isoelectronic with $\ce{N2}$. BO = (10 − 4)/2 = 3

$\ce{NO+}$ (14e): NO has 15e; removing one electron from the antibonding $\pi^*$ gives 14e → isoelectronic with $\ce{N2}$. BO = 3

$\boxed{\text{Answer: (a) 1, 3 and 3}}$