Official solution
S1 : (\[\sim\] p \[ \vee \] q) \[ \vee \] (q \[ \vee \] p) = (q \[ \vee \] \[\sim\] p) \[ \vee \] (q \[ \vee \] p)
S1 = q \[ \vee \] (\[\sim\] p \[ \vee \] p) = qvt = t = tautology
S2 : (p \[ \wedge \] \[\sim\] q) \[ \wedge \] (\[\sim\] p \[\vee\] q) = (p \[ \wedge \] \[\sim\] q) \[ \wedge \] \[\sim\] (p \[ \wedge \] \[\sim\] q) = C = fallacy