I will answer the first two parts of your question, i.e., $\sqrt N$ has a continued fraction of the form $[\lfloor \sqrt N \rfloor; \overline{a_1, a_2, \ldots, a_n, 2\lfloor \sqrt N \rfloor}]$.
We will assume that we already know that a quadratic irrational number has a periodic continued fraction.
First, we need the following theorem of Galois, the proof of which I take from Theorem 7.20 in I. Niven, H. S. Zuckerman, H. L. Montgomery, An Introduction to the Theory of Numbers, 5th ed., Wiley (New York), 1991:
Theorem of Galois. The continued fraction of the quadratic irrational number $\xi$ is purely periodic if $\xi > 1$ and $-1 < \xi' < 0$, where $\xi'$ is the conjugate of $\xi$.
Proof: Recall the inductive definition for expanding $\xi = \xi_0$ into a continued fraction:
$$a_i = \lfloor \xi_i \rfloor, \qquad \xi_{i+1} = \frac{1}{\xi_i - a_i}.$$
Invert and take conjugates of the latter equation to obtain
$$\frac{1}{\xi'_{i+1}} = \xi'_i - a_i.$$
Now $a_i \ge 1$. Hence, if $\xi'_i < 0$, then $1/\xi'_{i+1} < -1$, and we have $-1 < \xi'_{i+1} < 0$. Because $-1 < \xi'_0 < 0$ by hypothesis, we see by mathematical induction that $-1 < \xi'_i < 0$ is true for all $i$. Then by the previous equation,
$$0 < -\frac{1}{\xi'_{i+1}} - a_i < 1, \qquad a_i= \left\lfloor -\frac{1}{\xi'_{i+1}} \right\rfloor.$$
Because $\xi$ is a quadratic irrational number, it has a periodic continued fraction, so $\xi_j = \xi_k$ for some integers $j$ and $k$ with $0 < j < k$. Then we also have $\xi'_j = \xi'_k$, and
\begin{align}
a_{j-1} & = \left\lfloor -\frac{1}{\xi'_j} \right\rfloor = \left\lfloor -\frac{1}{\xi'_k}\right\rfloor = a_{k-1}\\
\xi_{j-1} & = a_{j-1} + \frac{1}{\xi_j} = a_{k-1} + \frac{1}{\xi_k} = \xi_{k-1}.
\end{align}
Hence, $\xi_j = \xi_k$ implies $\xi_{j-1} = \xi_{k-1}$. A $j$-fold iteration of that implication gives $\xi_0 = \xi_{k-j}$, and we have $\xi = \xi_0 = [\overline{a_0; a_1, \ldots, a_{k-j-1}}]$, which completes the proof of the theorem of Galois.
The rest of the answer comes from Proposition 32 of "Continued Fractions, Pell's Equation, and Transcendental Numbers" by Jeremy Booher:
Because
$$\frac{1}{\sqrt N - \lfloor \sqrt N \rfloor} > 1 \qquad \text{and} \qquad -1 < \frac{1}{-\sqrt N - \lfloor \sqrt N \rfloor} < 0,$$
the continued fraction of $\upsilon_1 = 1/(\sqrt N - \lfloor \sqrt N \rfloor)$ is purely periodic by the theorem of Galois, so $\upsilon_0 = \sqrt N$ has the form $[\lfloor \sqrt N \rfloor; \overline{a_1, a_2, \ldots, a_n, a_{n+1}}]$ where $n+1$ denotes the length of the shortest period in that expansion. (I am using the notation $\upsilon_0$ and $\upsilon_1$ because $\upsilon_0 = \sqrt N$ and $\upsilon_1 = 1/(\sqrt N - \lfloor \sqrt N \rfloor)$ are related by the inductive definition above for expanding $\upsilon_0$ into a continued fraction.) Also, $\sqrt N + \lfloor \sqrt N \rfloor = [2\lfloor \sqrt N \rfloor; \overline{a_1, a_2, \ldots, a_n, a_{n+1}}]$. Moreover,
$$\sqrt N + \lfloor \sqrt N \rfloor > 1, \qquad \text{and} \qquad
-1 < \lfloor \sqrt N \rfloor - \sqrt N < 0,$$
so the continued fraction of $\sqrt N + \lfloor \sqrt N \rfloor$ is purely periodic by the theorem of Galois, which means that the period (of length $n + 1$) starts at the beginning of the expansion. We therefore have more ways to express the continued fraction for $\sqrt N + \lfloor \sqrt N \rfloor$:
\begin{align}
\sqrt N + \lfloor \sqrt N \rfloor & = [2\lfloor \sqrt N \rfloor; \overline{a_1, a_2, \ldots, a_n, a_{n+1}}]\\
& = [\overline{2\lfloor \sqrt N \rfloor; a_1, a_2, \ldots, a_n}, a_{n+1}, \ldots]\\
& = [\overline{2\lfloor \sqrt N \rfloor; a_1, a_2, \ldots, a_n}].
\end{align}
We see from the last two continued fractions that $a_{n+1}$ begins the second period, so $a_{n+1}$ has the same value as the first integer of the first period, i.e., $a_{n+1} = 2\lfloor \sqrt N \rfloor$. Substituting for $a_{n+1}$ in the continued fraction for $\sqrt N$ concludes the proof:
$$\sqrt N = [\lfloor \sqrt N \rfloor; \overline{a_1, a_2, \ldots, a_n, 2\lfloor \sqrt N \rfloor}].$$
