The answer to Question 1 is negative. Take $K = {\mathbb Q}(\sqrt{-6})$
and let ${\mathfrak p} = (2,\sqrt{-6})$ denote the prime ideal above $2$.
The class number of $K$ is $2$, its maximal unramified (abelian) extension
is $L = {\mathbb Q}(\sqrt{-3},\sqrt{2})$. The ray class number formula shows that $h\{\mathfrak m\} = h \cdot \Phi(\mathfrak m)/(E:E^{(1)})$; here $\Phi$ is Euler's Phi function in $K$. In the present case, $E$ is generated by $-1$, hence the index in the denominator is $2$ for all ideals with norm $> 4$, and we simply have $h\{\mathfrak m\} = \Phi(\mathfrak m)$. We now compute the ray class numbers $h\{\mathfrak p^m\}$:
$$ \begin{array}{c|c}
m & h\{\mathfrak p^m\} \\ \hline
1 & 2 \\
2 & 4 \\
3 & 4 \\
4 & 8
\end{array} $$
Observe that $\Phi(\mathfrak p^m) = \Phi(\mathfrak p) \cdot N(\mathfrak p)^{m-1} = 2^{m-1}$ in our case.

This shows that the ray class field defined modulo $\mathfrak p^2 = (2)$ has conductor $(2)$ (because the ray class field defined modulo $\mathfrak p$ is strictly smaller) and that it contains the Hilbert class field of $K$ with conductor $(1)$.

In Question 2 there are conditions missing. The order $2^t$ is something I would expect for quadratic extensions.