Locknut & Majakdragon:
There are some inexpensive 1/2 inch solder-in water shut off valves that don't have packings or packing nuts. They rely entirely on a rubber O-ring to stop water leakage past the stem.
Rubber materials suffer from something called "compression set", which is where the rubber of a compressed O-ring (like the one between the stem and bonnet) will gradually flow into the shape it's compressed into. That is, you now have an O ring with an elliptical cross section, not a round cross section, even when it's not compressed into that shape.
When people see such flattened out O-rings, they presume the rubber on the ID of the O-ring has worn away, but that's not true since water shut off valves spend 99.9 percent of their time open with no movement of the spindle inside the O-ring. What's happened to cause that flattening out is "compression set"; the rubber has simply flowed into it's compressed shape. And, once that happens, it doesn't exert the outward pressure needed to stop leakage past it.
If this is an Emco compression stop, the fastest easiest fix is to replace the O-ring every 5 to 10 years. And, if it was me, I'd replace the rubber washer and the bibb screw at the same time.
And, if these are Emco compression stops, if you take the handle off, you should find that you can slip a 5/8 inch ID vinyl or rubber hose over the end of the bonnet nut. So, if the valve just won't stop leaking past the stem, slip a piece of 5/8" ID hose onto the bonnet nut, and put a container on the spot where the water drips out of the hose. Eventually, the valve should stop dripping.
The reason why the valve should eventually stop dripping is a matter of physics. For a fluid to leak past an O-ring, the pressure of the fluid has to be greater than the outward pressure exerted by the O-ring. Where the O-ring is compression set and isn't exerting any outward pressure of it's own, then theoretically, the OD of the O-ring is exactly equal to the ID of the bore that it's in. However, because the pressure on the opposite side of the O-ring in this case is atmospheric pressure, that O-ring is now being compressed by the fluid toward it's original shape. So, if the water in the piping is 45 psi, it will squeeze on the O-ring with a pressure of 45 psi, and the rubber of the O-ring will push back on the water with a pressure of 45 psi (otherwise the O-ring would move). However, rubber is not a solid, it's a very viscous liquid (which is why it compression set in the first place). So, if the pressure on the rubber is 45 psi, the pressure inside the rubber is 45 psi, and that pressure will be exerted outward in all directions, including against the bore the O-ring is in. So, the external pressure on the O-ring should result in exactly the same internal pressure inside the rubber, which will eventually cause it to seal against the bore with that same pressure. It all relies on the fact that rubber is a very viscous liquid and can flow in response to an applied force.
Locknut:
Where are those valves located? If they're readily accessible, it would be better to learn to solder and replace them. If they're not readily accessible, then it's a simple matter in a house to replace the washers and O-rings in them every half dozen years or so.