Your UPS is from 2009. I assume that the battery is also almost 7 years old as well. This may be the problem.
Lead acid battery 12 volt, 7Ah. Now UPS usually have a cutoff voltage where they stop draining the battery to prevent damage to the battery. For lead acid battery, they can be easily damaged when discharged past 50%. There are more expensive deep cycle lead acid batteries that can be discharged safely to 70-80%, but it is unknown which type your UPS has.
So let's assume that your UPS sets the cutoff point so that after 50% are used, it turns off. 12 volts X 3.5Ah = 42 Wh. Now add to this the fact that a 7 year old battery in a UPS will have suffered significant capacity loss. So now your useable energy will be even less than 42 Wh.
Your laptop is rated for 65 watts. So if it ran at the 65 watts for an hour it would require 65 Wh of energy and you have less than 42 Wh. I'll let you do the rest of the math, but you can see, even if you got 80% discharge, that low capacity battery coupled with its age - you will never get the several hours your looking for.
Adding to your dilemma, is that the inverter portion of your UPS is not perfect and you will lose at least 15% of your power to it. So 42 Wh is now down to 36 Wh.
It is possible that the UPS has a timer that shuts down after 15 minutes regardless of remaining power, and you could easily reset this by measuring the voltage of the battery immediately after it shuts down. If the voltage is around 11-11.5 volts then the battery probably ran out. If voltage is significantly higher, then a timer shut you down. You will need to measure the voltage right after shutdown, don't wait or the voltage is less reliable.
The best system is to simply have the charger constantly powering the battery. And the devices constantly drawing power from the battery at the same time. This is equivalent to a "full online" UPS. As there is zero transfer time. (cheap mains UPSes have a small drop in the mains output when they transfer the load from mains to battery during a power failure).
Check the output from your charger as mentioned above. As lots of 12V battery chargers have poor output voltage regulation. And are typically designed to charge a battery, and then be disconnected. They will often overcharge the battery if left connected to a battery 24/7. Get a regulated power supply that you can adjust the output to 13.5V or whatever the battery manufacturer specifies as the ideal "float" voltage for that battery.
Also consider how reliable your SEPIC power supplies are. Especially in relation to over voltage protection on their outputs. You don't want a faulty SEPIC inverter to destroy your connected devices from over voltage. Do a search for "shunt regulator" and "crowbar circuit" for examples of over voltage protection circuits.
And make sure that you put some fuses on the connections to your battery, as close to the positive terminal as practical. As lead acid batteries can provide very large currents if a short circuit occurs. The main purpose of the fuses in this case, is to stop a fire from occouring due to overloaded wires, if one of the SEPIC inverters or your battery charger fails short circuit.
Best Answer
The battery size will affect only its duration. If you get a 1Ah battery it will discharge in approximately one hour if you draw 1A, half an hour if you draw 2A and so on.
The second formula is good, keep in mind that cheaper UPS have a square wave output instead of a sine wave, some devices may be affected. The radio might keep some noise from the UPS and the fan might not work well since these electric motors usually expect a sine as input. You are safe to try them anyway, no risk of frying anything.
To calculate the battery duration you should proceed as follows:
$$I=\frac{P}{V}$$
where P is power, in W, and V is voltage rms, in V. This formula is valid only if the load is resistive, in the AC domain things can get a bit more complicated, your appliances should report maximum current consumption on a label somewhere, use that value. After that:
$$T = \frac{C}{I}$$
Where T is duration, in hours, C is battery capacity, in Ah or A times hours, and I is current, in A. The computed time is not the actual duration you would get but it's the upper theoretical limit.
If you live in europe and your radio is approximately a resistive load it would draw about 40mA, so a 1Ah battery would last no more than 23h15m.