Question about the proportions of primes p for which $ord_p(2)$ satisfies certain congruences let $p$ be a prime. 
I have observed numerically that the proportion of $p$ satisfying $ord_p(2)\equiv4[8]$ seems to be $1/3$. Why is it so? Is there a simple proof?
Moreover the proportion $c$ of prime $p$ satisfying  $ord_p(2)\equiv2[4]$ seems to be the same as the proportion of primes $p$ for which  $ord_p(2)\equiv1[2]$ or those $ord_p(2)$ which are odd. Again, why is it so?
$c\approx0.29..$  Who is that number?
 A: Recall that each residue class $1,3,5,7\pmod8$ contains $\frac14$ of the primes, asymptotically, by the prime number theorem for arithmetic progressions.
If $p\equiv3\pmod8$, then $p-1\equiv2\pmod4$, and $2$ is a quadratic nonresidue modulo $p$, which means that $ord_p(2)\equiv2\pmod4$ as well.
If $p\equiv7\pmod8$, then again $p-1\equiv2\pmod4$, but $2$ is now a quadratic residue modulo $p$, which means that $ord_p(2)\equiv1\pmod2$.
If $p\equiv5\pmod8$, then $p-1\equiv4\pmod8$, and $2$ is a quadratic nonresidue modulo $p$, which means that $ord_p(2)\equiv4\pmod8$ as well.
Finally, suppose that $p\equiv1\pmod8$, and let $k$ be the the exponent of $2$ in the factorization of $p-1$, so that $k\ge3$. Note that among these primes, $k=3$ for $\frac12$ of them, $k=4$ for $\frac14$ of them, $k=5$ for $\frac18$ of them, etc., again by the prime number theorem for arithmetic progressions.
Let $j$ be the the exponent of $2$ in the factorization of $ord_p(2)$. Since $2$ is a quadratic residue modulo $p$, we know that $j\le k-1$. We have $j=k-1$ if and only if $2$ is not a fourth power modulo $p$; $j=k-1$ if and only if $2$ is a fourth power but not an eighth power modulo $p$; $j=k-1$ if and only if $2$ is an eighth power but not a sixteenth power modulo $p$; and so on until $j=0$ if and only if $2$ is a $2^k$th power modulo $p$.
We can show (possibly requiring the assumption of a suitable generalized Riemann hypothesis) that among the primes $p\equiv1\pmod8$, $2$ is not a fourth power for $\frac12$ of them; $2$ is a fourth power but not an eighth power for $\frac14$ of them; $2$ is an eighth power but not a sixteenth power for $\frac18$ of them; and so on. (This is what we'd expect on the heuristic that $2$ is randomly distributed among the quadratic residues modulo $p$.)
Therefore the proportion of primes $p$ such that $ord_p(2)\equiv4\pmod8$ equals
$$
0+0+\frac14+\frac14\bigg( \frac12\cdot\frac12+\frac14\cdot\frac14+\frac18\cdot\frac18+\cdots \bigg) = \frac13.
$$
Similarly, the proportion of primes $p$ such that $ord_p(2)\equiv2\pmod4$ equals
$$
\frac14+0+0+\frac14\bigg( \frac12\cdot\frac14+\frac14\cdot\frac18+\frac18\cdot\frac1{16}+\cdots \bigg) = \frac7{24} = 0.291\bar6.
$$
A similar calculation gives $\frac7{24}$ for the proportion of primes $p$ such that $ord_p(2)$ is odd. Finally, the proportion of primes $p$ such that $ord_p(2)\equiv2^k\pmod{2^{k+1}}$ is $1/(3\cdot2^k)$.
