Official solution
(1) (p \[\to\] q) \[\vee\] (\[\sim\] q \[\to\] p)
= (\[\sim\] p \[\vee\] q) \[\vee\] (q \[\vee\] p)
= (\[\sim\] p \[\vee\] p) \[\vee\] q
= t \[\vee\] q = t
(2) (q \[\to\] p) \[\vee\] (\[\sim\] q \[\to\] p)
= (\[\sim\] q \[\vee\] p) \[\vee\] (q \[\vee\] p)
= (\[\sim\] q \[\vee\] q) \[\vee\] p
= t \[\vee\] p = t
(3) (p \[\to\] \[\sim\] q) \[\vee\] (\[\sim\] q \[\to\] p)
= (\[\sim\] p \[\vee\] \[\sim\] p) \[\vee\] (q \[\vee\] p)
= (\[\sim\] p \[\vee\] p) \[\vee\] (\[\sim\] q \[\vee\] q)
= t \[\vee\] t = t
(4) (\[\sim\] q \[\to\] q) \[\vee\] (\[\sim\] q \[\to\] p)
= (p \[\vee\] q) \[\vee\] (q \[\vee\] p)
= (p \[\vee\] p) \[\vee\] (q \[\vee\] p)
= p \[\vee\] q
Which is not a tautology.