1. Use bond order formula

According to MO theory,

where $N_b$ = number of electrons in bonding MOs and $N_a$ = number of electrons in antibonding MOs.

2. For $\mathrm{O}_2^{2-}$

First, $\mathrm{O}_2$ has 16 electrons, so $\mathrm{O}_2^{2-}$ has 18 electrons.

For oxygen and beyond, the MO filling order is:

In $\mathrm{O}_2$, bond order is 2. Adding two more electrons to form $\mathrm{O}_2^{2-}$ fills the two $\pi^*$ antibonding orbitals.

So bond order decreases by 1:

3. For $\mathrm{CO}$

Carbon has 6 electrons and oxygen has 8 electrons, so total electrons:

$\mathrm{CO}$ is isoelectronic with $\mathrm{N}_2$ (14 electrons).

For 14-electron species like $\mathrm{N}_2$ and $\mathrm{CO}$, bond order is:

So,

4. For $\mathrm{NO}^+$

Nitrogen has 7 electrons, oxygen has 8 electrons, and the positive charge means one electron is removed:

So $\mathrm{NO}^+$ is also isoelectronic with $\mathrm{N}_2$ and $\mathrm{CO}$.

Hence,

5. Final sequence

Thus, the bond orders for $\mathrm{O}_2^{2-}$, $\mathrm{CO}$ and $\mathrm{NO}^+$ respectively are:

So the correct option is:

A: $1, 3$ and $3$

6. Comparison with stored answer

Stored correct answer = A.

Our derived answer = A.

So they agree.