Start with this putting TTFF under p and TFTF under q 

p  q  ~p  ~q   ∼p∨q  (∼p∨q)→ ∼q
—————————————————————————————————
T  T
T  F
F  T
F  F

Under the ~p put the opposite of p. Since p has TTFF,
~p will have FFTT:

p  q  ~p  ~q  ∼p∨q  (∼p∨q)→ ∼q
————————————————————————————————
T  T   F
T  F   F
F  T   T
F  F   T

Under the ~q put the opposite of q. Since q has TFTF,
~q will have FTFT:

p  q  ~p  ~q  ∼p∨q  (∼p∨q)→ ∼q
————————————————————————————————
T  T   F   F
T  F   F   T
F  T   T   F 
F  F   T   T

We make ∼p∨q out of the columns ~p, q according to
the rule:

If there are F's on both sides of the disjunction
symbol ∨ the disjunction is F, otherwise it's T
Since ~p is FFTT and q is TFTF, then only the 2nd 
row has F under both ~q and p, so it gets F and the
other three get T, so under ∼p∨q we put TFTT

p  q  ~p  ~q  ∼p∨q  (∼p∨q)→ ∼q
————————————————————————————————
T  T   F   F    T 
T  F   F   T    F
F  T   T   F    T
F  F   T   T    T


We make (∼p∨q)→∼q out of the columns ∼p∨q and ∼q according to
the rule:

If there is a T on the left of the conditional 
symbol → and a T on the right side of the conditional
symbol, the conditional is F, otherwise it's T.
Since ∼p∨q is TFTT and ~q is FTFT, then the 1st 
and 3rd rows has F have this, so it gets F in rows
1 and 3
other two get T, so under (∼p∨q)→ ∼q
we put FTFT

p  q  ~p  ~q  ∼p∨q  (∼p∨q)→ ∼q
————————————————————————————————
T  T   F   F    T         F 
T  F   F   T    F         T
F  T   T   F    T         F
F  F   T   T    T         T

Edwin