Remember that the principal branch of $\log$ (which your book denotes by Log) is defined everywhere except for nonpositive real numbers.
For a: When your book says "the principal branch won't work", it means that the "obvious" answer of $\exp(\frac{1}{2} Log(z^2-1))$ isn't correct. This is a branch of $(z^2-1)^{1/2}$, but isn't defined everywhere on $|z|<1$. That's because at some points in $|z|<1$, the argument $z^2-1$ to Log is a nonpositive real number, where Log is not defined. Where does this happen? We have $z^2-1 \le 0$ iff $z^2 \le 1$. That naturally splits into 2 pieces: If $0 \le z^2 \le 1$, then $z$ will be in the real interval (-1,1) along the real axis, but if $z^2 \le 0$, then $z$ can be anywhere on the imaginary axis. So Log($z^2-1$) will fail on all those points.
However, if we use the function Log($1-z^2$) instead, then that is defined everywhere in the domain $|z|<1$. That's because $1-z^2$ cannot be a nonpositive real number if $|z|<1$ (check this for yourself). Now we can write $\exp(\frac{1}{2}Log(1-z^2))$, which is defined everywhere in $|z|<1$. That gives us a branch of $(1-z^2)^{1/2}$, which is close to, but not quite, what we want. We want a branch of $(z^2-1)^{1/2}$. So, we multiply a value of $(-1)^{1/2}$, which could be either $i$ or $-i$. Thus we arrive at the answer you show above: $i \exp(\frac{1}{2}Log(1-z^2))$ (multiplying by $-i$ instead of $i$ would also be correct; the negative of a branch of a square root is also a branch).
To check that this answer gives you a branch of $(z^2-1)^{1/2}$, you should square it and see that the result is $z^2-1$.
How might one come up with the idea to try Log($1-z^2$)? Here's one way. First, notice that our problem will be solved if we define a branch of $\log(z^2-1)$ in $|z|<1$, because once we do that we can write $\exp(\frac{1}{2} \log(z^2-1))$. However, we cannot choose the principal branch Log for log, for reasons we have already seen.
Let's look at the reason in a little more detail. The region $|z|<1$ is a unit disc centered at 0. What happens if we apply $z^2-1$ to it? If we square all points of the unit disc centered at 0, we still get the same unit disc. Then if we subtract 1, we get a unit disc centered at -1, which I will call $D$. Because $D$ contains nonpositive real numbers, the principal branch Log doesn't work. However, if we were to multiply $D$ by -1, we would get a unit disc centered at 1. That disc doesn't contain any nonpositive real numbers and therefore Log is okay. This suggests using $-(z^2-1) = 1-z^2$ as the argument for Log.
One more point: Negating the argument is a relatively harmless thing to do for logarithms because of the "identity" $$\log(-z) = \log(z) + \log(-1).$$ Here I put "identity" in quotes because this "identity", extrapolated from the true identity for real logarithms $\log(xy) = \log(x) + \log(y)$, isn't quite right for complex logarithms. You can't actually choose a branch for which the identity holds everywhere. Nevertheless, it suggests that, for the original problem, negating the argument $z^2-1$ to $1-z^2$ is a relatively harmless thing to do and we'll be able to recover by a relatively simple operation of adding $\log(-1)$. (In the actual solution, we multiplied by $(-1)^{1/2}$, which is what happens after you apply the exponential: $\exp(\frac{1}{2} \log(-1)) = (-1)^{1/2}$.)
I hope this helps you proceed similarly for the other 3 parts.
Best Answer
The trick with these radicals of polynomials is usually to factor them into linear factors and choose a branch of the logarithm for every factor such that cancellation occurs on some of the branch cuts. In this case the process is particularly simple and it suffices to choose one branch of the logarithm for both factors, which is the one that has the branch cut along the negative imaginary axis $$ -\frac{1}{2} \pi \le \arg \operatorname{Log} z < \frac{3}{2}\pi.$$ We will continue to use the lower case log for the branch with the cut along the negative real axis.
Now we define $$ f(z) = \exp(1/2 \operatorname{Log}(z-2i)) \exp(1/2 \operatorname{Log}(z+2i)).$$ The problem requires us to do two things: show that $f(z)$ is analytic in the complex plane slit between the two logarithmic singularities through the origin and verify that it agrees with $\sqrt{x^2+4}$ on the real line.
For the first part we certainly know that $f(z)$ is analytic in the complex plane slit from $2i$ to infinity along the negative imaginary axis. But we can show that it is in fact continuous across the slit below $-2i.$ To do this consider the point $z=-ti$ with $t>2$ and its two neighbors: $-ti-\epsilon$ in the left half plane and $-ti+\epsilon$ in the right half plane, with $\epsilon$ real and going to zero. For the left point we have $$ f(z) = \exp(1/2 \log(t+2) + 1/2 \times 3/2 \pi i) \exp(1/2 \log(t-2) + 1/2 \times 3/2 \pi i) \\= \sqrt{t^2-4} \exp(+3/2 \pi i).$$ For the right point we get $$ f(z) = \exp(1/2 \log(t+2) - 1/2 \times 1/2 \pi i) \exp(1/2 \log(t-2) - 1/2 \times 1/2 \pi i) \\= \sqrt{t^2-4} \exp(- 1/2 \pi i).$$ But since $\exp(- 1/2 \pi i) = \exp(+ 3/2 \pi i) = -i$ these two values are the same, so we have continuity across the cut. To get analyticity, apply Morera's theorem to a circle lying on the cut. Split it along a left half and a right half each consisting of an arc whose endpoints are connected by a line and integrate along both halves in counterclockwise direction. The two line segments cancel by continuity across the cut. The integral along the left segment is zero, as is the one along the right segment. Hence the integral along the circle lying on the cut is zero, and by Morera's theorem $f(z)$ is analytic there.
We still need to show that $f(z)$ agrees with $\sqrt{x^2+4}$ on the real axis. We will do the case $x>0.$ By Pythagoras the modulus of $x-2i$ is $\sqrt{x^2+4}$ as is the modulus of $x+2i.$ The point $x-2i$ forms some angle with the $x$ axis, call it $\theta$. The salient feature here is that $-1/2\pi < \theta < 0$ and that the point $x+2i$ forms that same angle with the $x$ axis, only positive, and there is no discontinuity in the angle because these two are entirely contained in the range of the argument of $\operatorname{Log} z.$ It follows that $$f(z) = \exp(1/2 \log\sqrt{x^2+4} + 1/2i\theta) \exp(1/2 \log\sqrt{x^2+4} - 1/2i\theta) \\=\exp(1/2 \log\sqrt{x^2+4}) \exp(1/2 \log\sqrt{x^2+4}) = \exp(\log\sqrt{x^2+4}) = \sqrt{x^2+4}.$$