When cell 2 is in its resting membrane potential (-70 mV), it means positively charged ions are more outside the cell than inside of the cell. Now, if you increase the concentration of extracellular potassium, it adds more positive charges outside. So, the difference (-70 mV) rises more (suppose -90 mV). This is hyper polarized state.
So, cell 2 would be less likely to fire an AP, as it is in a hyperpolarized state. Because hyper polarized state inhibits action potentials.
B) Sodium ions.
These channels permit influx of sodium ions that depolarizes postsynaptic membrane (that is membrane of cell 2).
C) If extracellular calcium was removed, during action potential less calcium would entered the presynaptic terminal. As calcium influx strengthen the synaptic transmission, less calcium in this case would result into less synaptic transmission. It will reduce the ability of cell 1 to excite cell 2.
Cell # 1 forms an excitatory synapse onto cell # 2 , when cell #1 fires a single action potential...