Quite easy.
You have found that ln (1 + tan x) = x - ½x² + ⅔x³ +...
So likewise, ln (1 + tan 2x) = 2x - ½(2x)² + ⅔(2x)³
= 2x - 2x² + 16/3 x³ + ...
Next, realise that sec² 2x / (1 + tan 2x)
= ½ (2sec² 2x / (1 + tan 2x) )
This is basically the derivative of ½ ln (1 + tan 2x) .
[Recall that d/dx tan (f(x)) = f'(x) sec² (f(x)) and d/dx ln (f(x)) = f'(x) / f(x) ]
Since ln (1 + tan 2x) = 2x - 2x² + 16/3 x³ + ... , then we just have to differentiate this and halve the result.

sec² 2x / (1 + tan 2x)
= ½ (2sec² 2x / (1 + tan 2x) )
= ½ d/dx ln (1 + tan 2x)
= ½ d/dx (2x - 2x² + 16/3 x³ + ...)
= ½ (2 - 4x² + 16x² + ...)
= 1 - 2x + 8x² + ...
We have found the first three terms.