Complex Numbers Geometry

I don't know how to use the hints you've been given, but you could solve these problems as follows.

First, note that the equation in (a) is equivalent to $\bar{a}z+a\bar{z}=2$ and the one in (b) to $(a+b)\bar{z}=2$ since $\bar{a}=1/a$ for any $a$ on the unit circle (I assume $a+b\ne0$).

If $a=e^{i\phi}$, then the tangent line is given by $x\cos\phi+y\sin\phi=1$. You can use this to prove (a) by plugging $z=x+iy$ and $a=\cos\phi+i\sin\phi$ into my version of (a) given above and taking the products.

For (b), assume $\bar{a}z+a\bar{z}=2$ and $\bar{b}z+b\bar{z}=2$ and take the sum $$\bar{a}z+a\bar{z}+\bar{b}z+b\bar{z}=(a+b)\bar{z}+(\bar{a}+\bar{b})z=4.$$ Note that $(a+b)\bar{z}$ is a real number since $a+b$ and $z$, which is the intersection point, have the same argument and the argument of $\bar{z}$ is the negative of that. Therefore, $(a+b)\bar{z}=2$.

Note that if $z$ is a point of the tangent line then: $$z=\lambda a i +a \quad (1)$$ and $$\bar{z}=-\lambda \bar{a} i + \bar{a} \quad (2)$$ where $\lambda$ is a real number.

Recall that if $a$ is in unit circle then: $$a \bar{a}=1$$ Let's calculate $z+ \bar{z} a^2$: $$z+ \bar{z} a^2=(\lambda a i +a)+(-\lambda \bar{a} i + \bar{a})a^2=2a$$ and we conclude that the statement (a) is true.

For statement (b) use twice statement (a) and solve the system, you will find that (b) is also true.