Schrödinger's cat bra-ket interpretation Let $\vert\text{#}\rangle$ be the vector state of the cat, $\vert1\rangle$ the "alive" state, and $\vert0\rangle$ the "dead" state.  Using the normalization condition $\langle \text{#}\vert\text{#}\rangle=1$:
\begin{equation}
\vert \text{#}\rangle=a\vert1\rangle+b\vert0\rangle
\end{equation}
becomes
\begin{equation}
\vert a\vert^{2}+a^{*}b\langle 1\vert0\rangle+b^{*}a\langle 0\vert1\rangle+\vert b \vert^{2}=1
\end{equation}
where $\vert a\vert^{2}$ is the probability of the cat being at the state $\vert1\rangle$ (alive).
The equation leads to $a=b=\frac{1}{\sqrt{2}}$. 
However, why is this? And how should $\langle 1\vert0\rangle$ and $\langle 0\vert1\rangle$ be interpreted?
 A: Following John Rennie's analogy, let the cat be a spin with up (alive) or down (dead) state. Note that you've expanded $|\#\rangle$ in terms of the orthonormal basis vectors:
$$|\#\rangle = a|1\rangle + b|0\rangle$$
By definition of orthogonality (and the idea that "alive" and "dead" are orthogonal), $\langle 0|1\rangle = \langle 1|0\rangle = 0$.
Now you're computing the probability that the cat is alive (hopefully it always is) or dead:
$$ |{\langle 1 |\#\rangle }|^2 = |a|^2, \quad |\langle 0 |\#\rangle|^2 = |b|^2 $$
To determine the range of values $a$ and $b$ can take, compute:
$$ |\langle\#|\#\rangle|^2 = |a|^2 + |b|^2 = 1$$
This is $1$ because $|\#\rangle$ is assumed to be normalised. So the range of values for $a$ and $b$ are not strictly $1/\sqrt{2}$ for both (for example $a = 1/\sqrt{5}, b = 2i/\sqrt{5}$ also satisfy the equation, where their absolute squares indicate the probability of being 'alive' or 'dead'), but $(|a|,|b|) \in S^1, \forall\;a, b \in \mathbb{C}$, where $S^1$ is the unit circle.
A: When you write:
$$ \vert \text{#}\rangle=a\vert1\rangle+b\vert0\rangle $$
you are assuming there exists an alive operator and that this operator has eigenstates $|1\rangle$ and $|0\rangle$ that you can use as a basis for writing the wavefunction of the cat. Neither of these assumptions seem reasonable so as it stands the question makes no sense.
However you could replace the cat by a spin that can be in a superposition of up and down states. In that case we are writing the $z$ component of the spin using the eigenstates of $\hat{L}_z$ as a basis, and we'd get the same equation:
$$ \vert \text{#}\rangle=a\vert1\rangle+b\vert0\rangle $$
where now our $|1\rangle$ and $|0\rangle$ states are well defined because they are the eigenstates of $\hat{L}_z$. And now it's clear that $\langle 0 | 1 \rangle$ and $\langle 1 | 0 \rangle$ are zero because eigenstates are orthogonal.
A: Your system has two mutually exclusive possible outcomes: dead cat $\vert 0\rangle$ or live cat $\vert 1\rangle$.  In analogy with spin up/down states, the $\vert 0\rangle$ and $\vert 1\rangle$ states are orthogonal, in the sense that if the cat is found to be alive (in the state $\vert 1\rangle$) then it is not dead - or more precisely it has 0% probability of being found dead: that's the meaning of $\vert \langle 0\vert 1\rangle\vert^2=0$.  Likewise, if the cat is found to be alive, it has 100% change of being alive: that's the meaning of $\vert \langle 1\vert 1\rangle\vert^2 =1$.
In this sense, a cat described by 
$$
\vert \#\rangle=a\vert 0\rangle + b\vert 1\rangle
$$
has probability $\vert \langle \#\vert 0\rangle\vert^2 = \vert  b\vert^2 $ of being found dead and $\vert \langle \#\vert 1\rangle\vert^2 = \vert  a\vert^2 $ of being found alive.  Note that the probabilities should sum to 1, i.e.
$\vert a\vert^2+\vert b\vert^2=1$.
Please note also that $a$ and $b$ can, in general, be complex numbers although of course their magnitude squared, which is a probability, is a real number, i.e. one could, in principle, have
$$
\vert \#\rangle =\frac{1}{\sqrt{3}}\vert 0\rangle + i\sqrt{\frac{2}{3}} 
\vert 1\rangle
$$
which would lead to measuring a dead cat $1/3$ of the time and a living cat $2/3$ of the time.
Finally, despite the surely bizarre nature of the original proposition, there are serious people working on this stuff: this 2015 article published in The Guardian reports on attempts to place microbes in a superposition of states.  
A: First of all, I do not think $a^2 + b^2 =1$ tells you that a = $\frac{1}{\sqrt 2}$. Basic trignometry. 
And to tell the truth, no one knows why $a^2$ is the probability(Or more precisely, the square of the coefficients). It's Born's Postulate.
<0|1> can be viewed as what is the probability density to find state |1> in |0>. And those two happens to be orthogonal if $|0>$ and |1> are a set of basis.
By the way, I think your question can be found in any quantum physics textbooks. Hope this helps you!
