To start, it is important that you understand that the methyl and hydroxy groups are activating only in the ortho and para positions. For example, if I were to substitute plain phenol, which contains an activating hydroxy group, which position would my electrophile (in this a bromine electrophile) add to? You can see that the intermediary sigma complex involves a positive charge. When the bromine attacks at the ortho or para (I did not include a diagram for the para attack, but it is similar to the ortho attack) positions, it is stabilized by an additional resonance structure provided by the oxygen lone pair. In the case of the meta attack, as you can see in the diagram below, this does not happen, and the intermediate is left with one less resonance structure.
Most activating groups will have the same directing effects as shown with the activating hydroxyl group above (para-ortho directors), so a methyl group, which contains no lone pair (and thus no additional resonance structures) will still stabilize the intermediate by donating electron density in much the same way as if an additional resonance structure was present.
Now getting back the para and meta cresol, the meta cresol contains two activating groups
Both of these activating groups activate the same positions, so this combination of activating effect will make the meta cresol more reactive than para-cresol. In para-cresol, the two groups do not work to activate the same positions, so, no matter which carbon is substituted, it will not be activated by both groups simultaneously like in meta-cresol.