Problem
Find Möbius transformations that send
$(i)$ the circle $|z|=2$ to $|z+1|=1$, and $-2$ to $0$, $0$ to $i$.
$(ii)$ the upper half-plane $Im(z)>0$ to $|z|<1$ and $\lambda$ to $0$ (where $Im(\lambda)>0$)
My attempt
For $(ii)$, I've tried to find $T(z)=\dfrac{az+b}{cz+d}$ that satisfied the conditions given. If $z_1,z_2,z_3$ are distinc points, and $w_1,w_2,w_3$ are also different points in the extended complex plane, then, there is a unique Möbius transformation that sends $z_i$ to $w_i$.
As they ask for $T(-2)=0$ and $T(0)=i$, there is only one more point $z \in \overline{\mathbb C} \setminus \{-2,0\}$ to define uniquely $T$. I've chosen $2 \in \{|z|=2|\}$ and I've sent it to $-2 \in \{|z+1|=1\}$.
So, the transformation constructed is $T(z)=\dfrac{z+2}{(-1+i)z-2i}$.
I've tried to show that for all $z \in \{|z|=2\}$, $T(z) \in \{|z+1|=1\}$.
In order to prove this, I've calculated $T(z)\overline{T(z)}=|T(z)|^2$
$$T(z)\overline{T(z)}=(\dfrac{z+2}{(-1+i)z-2i}+1)(\dfrac{\overline{z}+2}{(-1-i)\overline{z}+2i}+1)=(\dfrac{iz+2-2i}{(-1+i)z-2i})(\dfrac{-i\overline{z}+2+2i}{(-1-i)\overline{z}+2i})$$
Noting that $|z|=2$ and distributing, we have $$(\dfrac{iz+2-2i}{(-1+i)z-2i})(\dfrac{-i\overline{z}+2+2i}{(-1-i)\overline{z}+2i})=\dfrac{12+2Re(iz(2+2i)}{12+2Re(iz(-2+2i)}$$.
So, from here I cannot conclude $|T(z)|^2=1$, where have I committed a mistake? If my mistake doesn't come from arithmetics, how can I conveniently choose $z$ and define $T(z)$ for the remaining $z$ so that $T$ satisfies all the conditions?
The same goes for $(ii)$, I did something analogous and to $(ii)$ but arrived nowhere.
I would appreciate suggestions, thanks in advance.
While your approach is correct in theory, finding by hand an LFT from three distinct points can prove to be quite tedious. Without the aid of a computer, us humans normally categorize LFT's into groups and then prove properties about those groups for use later. Here are the two most common types of LFT's we should know by heart, Circle to Circle and Plane to Circle.
LFT 1. If $$f(z)=\frac{z-w}{1-\bar{w}z}$$ for some $w$, then $f(z)$ maps the unit circle to itself. (Try proving this by itself as an exercise.) Occasionally you made need a constant multiple of this instead, but not for the problems you gave.
LFT 2. Given two complex numbers $w_1$ and $w_2$, then $$f(z)=\frac{z-w_1}{z-w_2}$$ maps the half plane perpendicular to the midpoint line of $w_1$ and $w_2$ onto the unit disc. For example, in $$f(z)=\frac{z-(1+i)}{z-(-1-i)}$$ $w_1$ and $w_2$ have an midpoint of zero that is perpendicular to the line of $y=-x$ (draw this out). The right side of this line would be mapped onto the unit disc. This is so, because every $z$ to the right of $y=-x$ will be closer to the number $1+i$ than it will be to the number $-1-i$. Hence, $|f(z)|<1$ for $z$ to the right of $y=-x$ and by the same reasoning, $|f(z)|>1$ for $z$ to the left of $y=-x$. All the points on $y=-x$ are the same distance from $1+i$ and $-1-i$, hence the line $y=-x$ maps to the unit circle.
With these two tricks we can create almost any LFT we are looking for.
(Problem 1) Start with $$f(z)=\frac{z-w}{1-\bar{w}z}.$$ Compose and form $$g(z)=f(z/2)-1$$ so that $g(z)$ maps the circle $|z|=2$ to the circle $|z+1|=1$. Now pick $w=-1-i$, so that $-2$ maps to $0$ and $0$ maps to $i$. Simplify, $$g(z)=\frac{z/2-(-1-i)}{1-\overline{-1-i}(z/2)}-1=\frac{(i-1)(z+2)}{2(z+1+i)}.$$ It is not obvious from the formula itself that $g(z)$ has all the properties you desire. If you tried to proof $g(z)$ has the properties you want, you might be in for a terrible headache, namely because you would be trying to prove several interchanging arguments altogether. That is, why we prove it a piece at a time, first showing LFT 1 and then composing.
(Problem 2) Given $\lambda$ ($Im(\lambda)>0$) we pick a complex number on the other side of the upper half plane, namely $\overline{\lambda}$. The midpoint of $\lambda$ and $\overline{\lambda}$ is a number on the real line $\mathbb{R}$, hence the perpendicular line to the midpoint is the real line itself. Thus, $$f(x)=\frac{z-\lambda}{z-\overline{\lambda}}$$ maps the upper half plane to the unit circle as desired. Again, this follows immediately because any $z$ in the upper half plane is closer to $\lambda$ than to $\overline{\lambda}$, hence $|f(z)|<1$ for $z$ in the upper half plane.