Questions regarding dot product (possible textbook mistake) I am given the following exercise:

Show that $\Vert \overrightarrow{a} + \overrightarrow{b} \Vert = \Vert
 \overrightarrow{a} \Vert + \Vert \overrightarrow{b} \Vert $ if and
  only if $\overrightarrow{a}$ and $\overrightarrow{b}$ are parallel and
  point to the same direction. Also, show that $\Vert \overrightarrow{a} + \overrightarrow{b} \Vert = \Vert \overrightarrow{a} - \overrightarrow{b} \Vert$ if and only if $\overrightarrow{a} \cdot \overrightarrow{b} = 0$

Regarding the first question:
Firstly, i feel like the first question is wrong since $\Vert \overrightarrow{a} + \overrightarrow{b} \Vert = \Vert
 \overrightarrow{a} \Vert + \Vert \overrightarrow{b} \Vert $ for $\theta = 0$ or $\pi$ (so they must not necessarily point on the same direction). Is that correct?
if so, I proceeded the following way:
$$
\Vert \overrightarrow{a} + \overrightarrow{b} \Vert^2 = \Vert
 \overrightarrow{a} \Vert^2 + \Vert \overrightarrow{b} \Vert^2 \pm 2 \Vert \overrightarrow{a} \Vert \Vert \overrightarrow{b} \Vert \cos (\theta)\\
\Vert
 \overrightarrow{a} \Vert^2 + 2 \Vert \overrightarrow{a} \Vert \Vert \overrightarrow{b} \Vert +\Vert \overrightarrow{b} \Vert^2 = \Vert
 \overrightarrow{a} \Vert^2 + \Vert \overrightarrow{b} \Vert^2 \pm 2 \Vert \overrightarrow{a} \Vert \Vert \overrightarrow{b} \Vert \cos (\theta)\\
\cos(\theta) = \pm 1\\
\theta = 0 \text{ or } \theta = \pi
$$
Regarding the second question:
$$
\Vert
 \overrightarrow{a} \Vert^2 + \Vert \overrightarrow{b} \Vert^2 +2 \Vert \overrightarrow{a} \Vert \Vert \overrightarrow{b} \Vert \cos (\theta) = \Vert
 \overrightarrow{a} \Vert^2 + \Vert \overrightarrow{b} \Vert^2 -2 \Vert \overrightarrow{a} \Vert \Vert \overrightarrow{b} \Vert \cos (\theta)\\
- \cos(\theta) = \cos(\theta)\\
\cos(\theta) = 0\\
\theta = \pi/2
$$
Can someone help me with these?

Both of your questions can be answered using the fact that
$$
\lVert a\rVert^2 = a\cdot a
$$
So,
$$
\begin{align}
\lVert a + b\rVert^2 &= (a + b)\cdot (a + b) = \lVert a\rVert^2 + \lvert b\rVert^2 + 2a\cdot b.
\end{align}
$$
