Is $x^{\frac{3}{2}}$ a differentiable function? When we speak of differentiable function, do we mean $C^1$ function? 

I know $C^1$ means continuously differentiable.

Is $x^{\frac{3}{2}}$ a $C^1$ function or $C^2$ function? Its second derivative blow up at $x = 0$. Is it $C^1$?

Does Taylor series exist for $x^{\frac{3}{2}}$?
Yes. The domain of your function is $x\ge 0$ and the derivative $\frac 32\sqrt x$ is well defined in this domain.
Assume $x&gt;0$. Then
$$
f(x)=x^{3/2},\quad f'(x)=\frac32 \sqrt{x},\quad f''(x)=\frac3{4\sqrt{x}}.
$$ Observe that all these functions are continuous over $(0,\infty)$, that is $f$ is $C^1(0,\infty)$. In fact, $f$ is $C^\infty(0,\infty)$.