Part 2: Does the arithmetic mean of sides right triangles to the mean of their hypotenuse converge? A primitive Pythagorean triplet is a triplet $a^2 + b^2 = c^2$ be where $a,b,c$ have no common factors and is generated by $a = r^2 - s^2, b = 2rs, c = r^2 + s^2$ where $r > s \ge 1, \gcd(r,s) = 1$ and exactly one of the two numbers $r$ and $s$ is even. Clearly as $r$ increases, the number of primitive triplets formed for a given $r$ increases since the number of $s$ satisfying the above conditions increases.
Claim: Let $c_1,c_2,\ldots$ be the hypotenuse and $b_1,b_2,\ldots $ be the corresponding longer of the two orthogonal sides formed for Pythagorean triangles for all  $r \le x$ then as $x \to \infty$,

$$\frac{b_1 + b_2 + b_3 + \cdots}{c_1 + c_2 + c_3 + \cdots} = \sqrt{2} - \frac{1}{2}$$

Can this claim be proved or disproved?
The difference between this question and the related question: Part 1: Does the arithmetic mean of sides right triangles to the mean of their hypotenuse converge? is that here the triangles are in sequenced in ascending order of $r$ and $s$ where as in the related question, they are sequenced in ascending order of the hypotenuse and depending on the choice of sequencing, the limiting value differs.
 A: Your claim is right. To see why, 
consider the quantities
$$
B_r=\sum_{\text{constant }r} b_{r,s},
\quad
C_r=\sum_{\text{constant }r} c_{r,s}.
$$
I will show that $B_r/C_r$ tends to $\sqrt2-1/2$ for $r\to\infty$.
Given any integer $r\ge2$ the possible values for $s$ are $r-1$, $r-3$, $r-5$, ... as long as $s>0$. We can summarise that as follows: 
$$
s=r-2k+1,\quad\text{where}\quad 1\le k\le \lfloor{r/2}\rfloor.
$$
We must take into account that $b_{r,s}$ can be given by two different expressions:
$$
b_{r,s}=\max(r^2-s^2, 2rs)=
\cases{
2r(r-2k+1) & for $1\le k<N(r)$,\\
r^2-(r-2k+1)^2 & for $N(r)<k\le\lfloor{r/2}\rfloor$,\\
}
$$
where $N(r)={2-\sqrt2\over2}r+{1\over2}$ is the value of $k$ for which $r^2-s^2=2rs$.
We may then compute $B_k$ as follows:
$$
B_k=\sum_{k=1}^{\lfloor{N(r)}\rfloor}2r(r-2k+1)+
\sum_{\lfloor{N(r)}\rfloor+1}^{\lfloor{r/2}\rfloor}r^2-(r-2k+1)^2.
$$
To keep the computation as simple as possible, considering that we want to find the limit $B_r/C_r$ for $r\to\infty$, we can keep only the leading terms in $r$ in the above expression. We can then substitute $\lfloor{r/2}\rfloor$ with $r/2$ and $\lfloor{N(r)}\rfloor$ with ${2-\sqrt2\over2}r$; moreover, we can discard $1$ in $r-2k+1$. This leads to:
$$
B_k\approx
\sum_{k=1}^{r(2-\sqrt2)/2}2r(r-2k)+
\sum_{r(2-\sqrt2)/2+1}^{r/2}r^2-(r-2k)^2=
{2\sqrt2-1\over3}r^3.
$$
We can repeat the same computation for $C_k$, obtaining:
$$
C_k=
\sum_{k=1}^{\lfloor{r/2}\rfloor} r^2+(r-2k+1)^2\approx
\sum_{k=1}^{r/2}r^2+(r-2k)^2=
{2\over3}r^3.
$$
Hence we obtain:
$$
\lim_{r\to\infty}{B_r\over C_r}=
\sqrt2-{1\over2}.
$$
From there, it is not difficult to show that
$$
{\sum_{r=2}^{\infty}B_r\over \sum_{r=2}^{\infty}C_r}=
\sqrt2-{1\over2},
$$
because both $B_r$ and $C_r$ asymptotically grow as $r^3$.
EDIT.
The same reasoning can be repeated for the shorter leg:
$$
A_r=\sum_{\text{constant }r} a_{r,s}=
\sum_{k=1}^{\lfloor{N(r)}\rfloor}r^2-(r-2k+1)^2+
\sum_{\lfloor{N(r)}\rfloor+1}^{\lfloor{r/2}\rfloor}2r(r-2k+1)
\approx{7-4\sqrt2\over6}r^3,
$$
leading to
$$
{\sum_{r=2}^{\infty}A_r\over \sum_{r=2}^{\infty}C_r}=
{7\over4}-\sqrt2.
$$
