Branches of analytic functions 
Find a branch of $\log(z^2+1)$ that is analytic at $z=0$ and takes the
  value of $2\pi i$ there. Also, determine a branch of $\log(z^2+2z+3)$ that is analytic at $z=-1$.

If I plug in $z=0$ and $z=-1$ to there respective functions then I get $\log(1)$ and $\log(2)$ but then what do I have to do to find a branch? 

Define a branch of $(z^2-1)^{1/2}$ that is analytic in the exterior of
  the unit circle $|z|>1$.

If I transform this function into $e^{(1/2)Log(z^2-1)}$, why will the principal branch not work here? 
 A: For your first question: To find a branch of $\log(z^2+1)$ means to find an analytic function $f$ such that $\exp(f(z)) = z^2+1$.  Here you haven't specified what the domain of $f$ should be, but presumably it should be as large as possible (and of course containing 0, which is stated in the problem).  
Hint: Factor $z^2+1$ and use the standard properties of logarithms from the real numbers.  Even though these properties don't hold in all cases for complex logarithms, they will help you to guess the answer, which you can then prove directly by exponentiating, as I described in the answer to your other question.
For your second question: The principal branch does not work because for some values of $z$ in $|z|>1$ (for example $z=2i$), the value of $z^2-1$ will be a negative real number, where $Log$ is not defined.
A: Let me give a simpler function so you can see what's going on. Let's consider the function $f(z)=\ln(z)$. We would like to find a branch where $\ln(z)$ is analytic at $z=1$ and has a value $2\pi i$ there. Now, you know that
$$ \ln(z)=\ln|z|+i(\theta+2k\pi) \longrightarrow  (1). $$
We want to find the branch of the function that has value $2\pi i$ at 
$$z=1 \implies z=e^{0i} \implies \theta = 0 . $$
Substituting $\theta =0 $ in $(1)$, $z=1$, and choosing the right branch, in our case $k=1$, we have
$$ \ln(1)= 0 + i(0+2\pi)=2\pi i. $$
I hope this example makes it clear for you. 
