What is happening with voltage on capacitor?
Capacitor is fully charged and switch is closed


I'm not sure what you mean be this.  Before the switch is closed, there is no path for current through the capacitor and thus, the voltage on the capacitor must be specified as part of the problem, i.e., the voltage across the capacitor, just before the switch is closed, is an initial condition that must be specified.  Stating that the capacitor is fully charged is not meaningful here without further context.

It is true that if the initial voltage on the capacitor is \$v_C(0) = IR_1\$, then the capacitor voltage will remain unchanged since, in that case, there will be no current through \$R_2\$ to change the voltage across the capacitor.  Put another way, the initial capacitor voltage equals the DC steady state capacitor voltage for the circuit with the switch closed.

If this is what you mean by 'fully charged', then the voltage across the capacitor is constant and equal to \$IR_1\$ and not \$0\mathrm{V}\$.

However, if the initial voltage is \$v_C(0) = 0\mathrm{V}\$, the capacitor is, by definition, uncharged and thus, after the switch is closed, will charge towards its steady state value of \$IR_1\$ with time constant \$\tau = (R_1 + R_2)C\$