Take the integer restriction off k and you get a good hint.
Maximize[{\[Omega], -(Pi/6) + (k*Pi)/\[Omega] <= Pi, -(Pi/6) + ((k + 1)*Pi)/\[Omega] >= (4*Pi)/3, \[Omega] > 0}, {\[Omega]}]
(*{Piecewise[{{(2*(k + 1))/3, Inequality[-1, Less, k, LessEqual, 7/2]}},-Infinity], {\[Omega] -> Piecewise[{{(2*(k + 1))/3, Inequality[-1, Less, k, LessEqual,
7/2]}}, Indeterminate]}}*)
From this you know that -1 < k <= 7/2
So make a table
Table[{k, Maximize[{\[Omega], -(\[Pi]/6) + (k \[Pi])/\[Omega] <= \[Pi], -(\[Pi]/6) + ((k + 1) \[Pi])/\[Omega] >= (4 \[Pi])/3, \[Omega] > 0},{\[Omega]}]}, {k, 0, 5}]
{{0, {2/3, {\[Omega] -> 2/3}}}, {1, {4/3, {\[Omega] -> 4/3}}}, {2, {2,{\[Omega] -> 2}}}, {3, {8/3, {\[Omega] -> 8/3}}}, {4, {-\[Infinity],{\[Omega] -> Indeterminate}}}, {5, {-\[Infinity], {\[Omega] -> Indeterminate}}}}
And now we know the maximum occurs at k = 3
if k must be an integer. Otherwise it occurs at k = 7/2
.