Why is there a negative in front of the phase response equation for this complex exponential?
For the magnitude, the denominator term will be $$1-0.9e^{-j\omega}=1-0.9(cos\omega - jsin\omega) = 1-0.9cos\omega+0.9jsin\omega$$

Hence the plus sign.

For the phase, if you multiply the numerator and denominator with the complex conjugate,

$$H(e^{j\omega}) = \frac {1-0.9cos\omega-0.9jsin\omega}{(1-0.9cos\omega+0.9jsin\omega)(1-0.9cos\omega-0.9jsin\omega)}$$

The denominator will become real.
Therefore,
$$\angle H(e^{j\omega}) = arctan(\frac {-0.9 sin\omega}{1-0.9cos\omega})$$
Taking the minus sign outside is how you get the negative in the phase.