The idea here is that the gas produced by a chemical reaction is often collected over water, which implies that it is mixed with water vapor.
In order to find the partial pressure of the dry gas, which is the gas produced by the reaction without the presence of the water vapor, you must use the partial pressure of water vapor at the temperature at which the gas is being collected.
In other words, you must use Dalton's law of Partial Pressures, which states that the partial pressure of a gaseous mixture is equal to the sum of the partial pressure of its gaseous components.
In your case, you will have
#P_"mixture" = P_"dry gas" + P_"water vapor"#
which will get you
#P_"dry gas" = P_"mixture" - P_"water vapor"#
Now, convert the temperature to degrees Celsius
#t[""^@"C"] = "308 K" - "273.15 K" = 34.85^@"C"#
Now look up the partial pressure of water vapor at this temperature
You'll find it listed as
#P_"water vapor" = "41.72 mmHg"#
which is equivalent to
#P_"water vapor" = "41.72 torr"#
You can now say that the partial pressure of the dry gas is equal to
#P_"dry gas" = "742 torr" - "41.72 torr"#
#color(darkgreen)(ul(color(black)(P_"dry gas" = "700. torr")))#
The answer is rounded to three sig figs.