In a particular month of some year, there are three Mondays which have even dates. On which day of the week does the $15\text{th}$ of that month fall?

Days of the week repeat every $7$ days. If a day falls on date $d$, the subsequent occurrences will be on $d+7,d+14,d+21,$ and $d+28$.
Notice that adding $7$ changes an even date to an odd date, and an odd date to an even date. Thus, the dates for Mondays in any month alternate between even and odd numbers.

To have exactly three Mondays with even dates, there must be a total of five Mondays in that month, arranged in the order:

$$\text{Even, Odd, Even, Odd, Even}$$

Let the first Monday be on an even date $d$.

* $1\text{st}$ Monday $=d$ (Even)
* $2\text{nd}$ Monday $=d+7$ (Odd)
* $3\text{rd}$ Monday $=d+14$ (Even)
* $4\text{th}$ Monday $=d+21$ (Odd)
* $5\text{th}$ Monday $=d+28$ (Even)

Since a month has at most $31$ days, the maximum value for the $5\text{th}$ Monday is $31$:

$$d+28\le 31⟹d\le 3$$

Since $d$ must be a positive even integer, the only possible value is $d=2$.
Substituting $d=2$, the dates of the Mondays are $2,9,16,23,$ and $30$.

Since the $16\text{th}$ of the month is a Monday, the $15\text{th}$ of the month must be a day before Monday, which is Sunday.