Electronic – Calculate minimum voltage to turn off/on solid state relay

This type of opto-triac is mostly used in mains voltage applications. Due to the limited current capabilities it's often used as a driver for a triac which is the actual switching device. Your requirements are modest, so you won't need that, and you can use the opto-triac to switch your load directly. The opto-triac is a cheaper solution than an electromechanical relay then, so at first sight looks like a better choice.
An important difference between electronic and electromechanical switches, however, is that the latter have a very low on-resistance, while the former always will have a voltage drop when switched on. That's the on-state voltage mentioned in the datasheet. This can be up to 3V, which in a 230V application won't matter much, but if your supply voltage is only 24V AC that's more than 10%. Your load will probably work at 21V, but you'll have to check it.

Repetitive peak off-state current is the leakage current when the triac is switched off. 2\$\mu\$A is a safe value.

Holding current is the minimum load current the triac needs to remain on when the gate is no longer driven. For an average triac your 20mA may be a bit low, but again the opto-triac's 3.5mA is a safe value. (Besides, the gate will be continuously driven, so it's a moot point. It is important in four-component dimmers, where the diac gives a pulse to switch on the triac, after which the triac is on its own.)

Then there is the minimum trigger current. That's the minimum current you have to supply to the LED to switch the triac on, and we'll have to calculate the series resistor accordingly.
Where did you get that 38\$\Omega\$ resistor value? You need figures 3 and 4 to calculate the value for the LED resistor. Figure 4 shows that 10mA is a safe value, and figure 3 shows that at 10mA the LED voltage will be maximum 1.3V. So \$R=\frac{3.3V - 1.3V}{10mA}=200\Omega\$ maximum. Your 38\$\Omega\$ would result in more than 50mA, which is not only more than Absolute Maximum Ratings (page 4), but also more than your microcontroller will be able to supply. So don't exaggerate, and pick a 180 \$\Omega\$ resistor. At lower resistances the current may become too much for your microcontroller's output. If you want more current through the LED (no more than 20mA, never use the Absolute Maximum Ratings!) you may want to use a transistor. Since you'd need a lot of them, consider a driver IC like an ULN2803.

In conclusion I think this opto-triac is a good choice. Alternatively, you may have a look at the MOCxxx series, for instance the MOC3012 needs only half of the LED current, which your microcontroller would appreciate. It doesn't give a nominal value for triac current directly, but from maximum power dissipation (300mW) we can derive that this should be 100mA. (It says peak repetitive surge current is 1A, 120pps, 1ms pulse width.)

This sounds like a good application for regular old mechanical relays. They won't care about the direction of current, won't have a offset voltage, and their off leakage is zero in a practical sense.

Relays that can switch house power are common and available in a variety of coil voltages. Since you say you want to switch "lights", it seems like the current will be low. Relays that can do this will be small and available in coils meant to be driven from 5 V. This means you can run them from the same power supply as the processor doing the controlling, but not that the relays can be run directly from a digital output. You will need a transistor controlled by the digital output that in turn controls the larger relay coil current. Don't forget the reverse diode accross the relay coil, else you'll end up with fried transistors.