Is there a way to calculate bond energies using computational chemistry?
Is there a way to calculate bond energies using GAMESS?


Yes. You would normally model the bound and unbound states

$$\ce{ML_{n - 1} + L -&gt; ML_{n}}$$

The definition of bond energy is based on the enthalpy of reaction in the gas phase (thanks DSVA!), so you would calculate it as follows:

$$\Delta H^\circ = H_{\ce{ML_{n}}} - (H_{\ce{ML_{n - 1}}} + H_{\ce{L}})\text{.}$$



While not strictly bond energy by definition, a value for the reaction Gibb's free energy might also be useful in some contexts:

$$\Delta G^\circ = G_{\ce{ML_{n}}} - (G_{\ce{ML_{n - 1}}} + G_{\ce{L}})\text{.}$$

This takes entropy into account.
For the calculation of Gibb's free energies, most codes calculate the molecular partition function from translational, rotational and vibrational contributions and get enthalpy and entropy from it.
Due to the vibrational contributions, you'll need to diagonalize the Hessian matrix of the system. This will give you an association (minus dissociation) energy value.
(You might take solvation into account through explicit/implicit solvation, use decent level of theory, validate/benchmark, etc.)
There are a lot of approximations and considerations in the process. More can be found here (some equations) and here (more practical).

Observe that

$$K = e^{- \frac{\Delta G^\circ}{R T}}$$

is the equilibrium constant associated with the process above, which is called the stability constant, formation constant or binding constant.
This is widely used to "measure the strength of the interaction between the reagents that come together to form a complex" (source).
Bear in mind that $\Delta G^\circ$ goes inside an exponential, i.e., small errors in $\Delta G^\circ$ become (exponentially) large errors in $K$, so $\Delta G^\circ$ might be more useful in a computational context than $K$ itself.




  Is there a way to determine if the complexes would form?


This is complicated.

Your reaction might be too slow though, i.e., actual formation might depend on transition states. Furthermore, there might be faster processes in competition that you might miss.