You don't specify an initial state. From your result for $A$, I infer that you're interested in the expected score in equilibrium (which is $\frac14$ per draw in $A$), not starting from a full deck (which would be slightly different).
In the general case, in a shuffled deck the $i+j$ non-reshuffle cards are equidistributed over the $k+1$ intervals formed by the $k$ reshuffle cards. Thus, on average we draw $\frac{i+j}{k+1}$ non-reshuffle cards before drawing a reshuffle card, and the expected score from these is $\frac i{k+1}$. Thus the expected score per draw is
$$
\frac{\frac i{k+1}}{\frac{i+j}{k+1}+1}=\frac i{i+j+k+1}
$$
(where $+1$ counts the reshuffle card). In your two specific cases, this yields
$$
\frac1{1+0+2+1}=\frac14
$$
and
$$
\frac1{1+10+2+1}=\frac1{14}\;,
$$
respectively.
Note that interestingly it doesn't matter how many of the non-scoring cards are zeros or reshuffles. As long as there's at least one reshuffle card, the expected score per draw is simply $\frac i{n+1}$, where $n$ is the total number of cards, compared to $\frac in$ if you loop through a deck without reshuffle cards.
Edit in response to a comment:
The ratio $\frac{\frac i{k+1}}{\frac{i+j}{k+1}+1}$ comes about as follows: Each time we draw from a full, shuffled deck up to and including a reshuffle card, the expected score is $\frac i{k+1}$ and the expected number of draws is $\frac{i+j}{k+1}+1$. For one such run, the expectation of the score per draw is hard to determine; it isn't simply the ratio of the two expectations. For instance, in your first example, we score $0/1$ with probability $\frac23$ and $1/2$ with probability $\frac13$, so the expectation of the score per draw is $\frac23\cdot0+\frac13\cdot\frac12=\frac16\ne\frac14$. However, if we perform a very large number $m$ of such runs, the number of draws will be close to $m$ times the expected number of draws per run, and the score will be close to $m$ times the expected score per run, so for all the runs together, the expectation of the score per draw is simply the ratio between the expected score per run and the expected number of draws per run.
I do not have a lot of time now, so let me look at the circle case only. The assumption of one fixed point was a wise shortcut. I might exploit the symmetry to suggest another here. Let $\theta$ be the central angle subtended by the two points. It is only necessary to integrate from 0 to $\pi$. Let $r$ be the circle radius.
$\frac{2r}{\pi} \int_{0}^{\pi} \sin \frac{\theta}{2} d\theta = \frac{4r}{\pi}$
Edit:
Having more time now, I might look at the sphere. I am not exactly following your integral, but we are reaching the same bottom line, so that looks good. Before starting the sphere, I want to come clean on my circle. There was a typo in the integral, but remarkably it led to the same result. I have since corrected it.
Let $\varphi$ be the angle subtended by the fixed point and the variable point on the sphere. Again use radius $r$. The locus of the variable points defining that angle is a circle with radius $r\sin \varphi$, so its circumference is $2\pi r\sin \varphi$. Let it mark a strip of width $rd\varphi$, giving it area $2\pi r^2 \sin \varphi d\varphi$. The probability of the point falling on that strip is that area divided by the total surface area of the sphere.
probability = $\frac{2\pi r^2 \sin \varphi d\varphi}{4\pi r^2} = \frac{1}{2} \sin \varphi d\varphi$
length of chord = $2r\sin \frac{\varphi}{2}$
expectation = $r \int_{0}^{\pi} \sin \varphi \sin \frac{\varphi}{2} d\varphi$
= $2r \int_{0}^{\pi} \sin^2 \frac{\varphi}{2}\cos \frac{\varphi}{2} d\varphi$
= $\frac{4r}{3} [\sin^3 \frac{\varphi}{2}]_{0}^{\pi}$
= $\frac {4r}{3}$
I tend to use geometry as far as it will carry me. In this case, that served to simplify both integrals. Not everyone responds to that. See what you think.
Best Answer
The asymptotics is $1/\sqrt6=0.40824829$.
To see this, consider i.i.d. random variables $X_i$ and $Y_i$ uniform on $[0,1]$ and write the quantity to be computed as $I_k=\mathrm E\left(\sqrt{Z_k}\right)$ with $$ Z_k=\frac1k\sum_{i=1}^k(X_i-Y_i)^2. $$ By the strong law of large numbers for i.i.d. bounded random variables, when $k\to\infty$, $Z_k\to z$ almost surely and in every $L^p$, with $z=\mathrm E(Z_1)$. In particular, $I_k\to \sqrt{z}$. Numerically, $$ z=\iint_{[0,1]^2}(x-y)^2\mathrm{d}x\mathrm{d}y=2\int_0^1x^2\mathrm{d}x-2\left(\int_0^1x\mathrm{d}x\right)^2=2\frac13-2\left(\frac12\right)^2=\frac16. $$
Edit (This answers a different question asked by the OP in the comments.)
Consider the maximum of $n\gg1$ independent copies of $kZ_k$ with $k\gg1$ and call $M_{n,k}$ its square root. A heuristics to estimate the typical behaviour of $M_{n,k}$ is as follows.
By the central limit theorem (and in a loose sense), $Z_k\approx z+N\sqrt{v/k}$ where $v$ is the variance of $Z_1$ and $N$ is a standard gaussian random variable. In particular, for every given nonnegative $s$, $$ \mathrm P\left(Z_k\ge z+s\right)\approx\mathrm P\left(N^2\ge ks^2/v\right). $$ Furthermore, the typical size of $M_{n,k}^2$ is $z+s$ where $s$ solves $\mathrm P(Z_k\ge z+s)\approx1/n$. Choose $q(n)$ such that $\mathrm P(N\ge q(n))=1/n$, that is, $q(n)$ is a so-called quantile of the standard gaussian distribution. Then, the typical size of $M_{n,k}^2$ is $k(z+s)$ where $s$ solves $ks^2/v=q(n)^2$. Finally, $$ M_{n,k}\approx \sqrt{kz+q(n)\sqrt{kv}}. $$ Numerically, $z=1/6$, $v=7/180$, and you are interested in $k=1'000$. For $n=10'000$, $q(n)=3.719$ yields a typical size $M_{n,k}\approx13.78$ and for $n=100'000$, $q(n)=4.265$ hence $M_{n,k}\approx13.90$ (these should be compared to the values you observed).
To make rigorous such estimates and to understand why, in a way, $M_{n,k}$ concentrates around the typical value we computed above, see here.