Guidance requested for vector dot product question.

I don't know what your son is supposed to know about the dot product... which is critical for the answer.

However if he knows that dot product is distributive vs. addition then $$a \cdot (b-a) = a \cdot b - a \cdot a=0.$$

Therefore 2. is clear and 4. also as $a \cdot a = \vert a \vert^2$.

Your figure is an example of how $a\cdot(b - a)$ could be true. Note that $b$ is a different vector from $a$ in that figure, so it is an example where $a = b$ is false.

For the rest, in general, if vectors $x$ and $y$ are perpendicular the $x \cdot y = 0.$ This is often used as part of the definition of what it means for vectors to be perpendicular.

But $a\cdot (b - a) = a\cdot b - a \cdot a$ (distribution of dot product over addition), so $a\cdot (b - a) = 0$ implies $a\cdot b - a \cdot a = 0$, which implies $a\cdot b = a \cdot a.$ Moreover, in general for any vector $a,$ we always have $|a|^2 = a\cdot a.$