Why does $\sqrt 5$ occur in manageable situations of these scenarios? 
*

*Banach-Mazur distance between $P_5$ and $P_3$ is $d(P_5,P_3)=1+\frac{\sqrt5}2$ where $P_n$ is regular polygon in $n$ sides https://ieeexplore.ieee.org/stamp/stamp.jsp?tp=&arnumber=7968198&tag=1. 

*Shannon zero error capacity of Pentagon is $\sqrt 5$ http://web.cs.elte.hu/~lovasz/scans/theta.pdf.

*Lovasz Theta and regular odd sided polygon agree and are algebraic for Pentagon https://en.wikipedia.org/wiki/Lov%C3%A1sz_number (similar to 2. but this resemblance is on tightness of semi-definite programming and algebraicity).

*$5$ is minimum sum of squares of two distinct natural numbers and also appears in Hurwitz theorem https://en.wikipedia.org/wiki/Hurwitz%27s_theorem_(number_theory) and seems related to geometry https://www.jstor.org/stable/pdf/2302799.pdf.

Does the presence of $\sqrt 5$ somehow make certain things easier by inducing spectacular constraints based on symmetries and in particular interest to me why is it difficult to prove 2. and 3. for any odd number above $5$?

Are there other scenarios where $\sqrt 5$ appeared and a seemingly hard general situation becomes tame with situation at hand?
Perhaps this is coincidence however it seemed hidden reason is plausible.
The answer so far does not address the problem.
 A: The geometric reason for the ubiquity of $\sqrt 5$ in problems involving a pentagon is that it is the diagonal of a  $1\times 2$ rectangle (a "half-square"). This links $\sqrt 5$ to the construction of a pentagon from its side, which may be at the origin of the first three geometric observations in the OP.
The fourth observation on Hurwitz theorem is not geometric, but it does involve the golden ratio, which is the side-to-diagonal ratio in a regular pentagon and in this way brings us back to $\sqrt 5$.
A: $\def\QQ{\mathbb{Q}}$Three of these four examples involve $5$-fold symmetry. If $\zeta_5$ is a primitive $5$-th root of unity, then $\zeta_5 + \zeta_5^{-1} = \tfrac{1 \pm \sqrt{5}}{2}$. So any computations with $5$-fold symmetry are likely to include square roots of $5$.
In the same way, if $p$ is any prime which is $1 \bmod 4$, then $\sqrt{p} \in \QQ(\zeta_p)$.
To connect this to the fact that $p = a^2+b^2$, I have to work a little harder, but I can say something. Let $p \equiv 1 \bmod 4$ be prime, and $L = \QQ(i, \zeta_p)$. The Galois group of $L/\mathbb{Q}$ is $C_2 \times C_{p-1}$, so it has a quotient $C_2 \times C_4$. (I write $C_n$ for the cyclic group of order $n$.) Let $K$ be the corresponding $C_2 \times C_4$ extension of $\QQ$.
Then $K/\QQ(i)$ is a $C_4$-extension so, by Kummer's theorem, $K = \QQ(i)(\sqrt[4]{\alpha})$ for some $\alpha \in \QQ(i)$. One can show that one can take $\alpha$ of the form $(a+bi)^3 (a-bi)$ for some $a+b i \in \QQ(i)$. Then $\sqrt{\alpha} = (a+bi) \sqrt{a^2+b^2}$ is in $K$ and thus $\sqrt{a^2+b^2}$ is in $K$. If you trace through the Galois theory, the element $\sqrt{a^2+b^2}$ is in $\QQ(\zeta_p)$. So we see that, if $\QQ(\sqrt{D})$ is the unique quadratic subfield of $\QQ(\zeta_p)$, then $D$ is of the form $a^2+b^2$. Of course, this is not at all the easiest way to show that a prime which is $1 \bmod 4$ is a sum of two squares!
A: I'm not sure whether $\sqrt5$ is anywhere special, as the same relation holds for primes $p$ with form $4k+1$.
(2) (3) Let $G$ be a Paley graph with vertices in $\mathbb F_p$. $G$ is self-complementary and vertex-transitive, so $\vartheta (G)\vartheta ({\bar  {G}})=p$, and it follows that $\vartheta (G)=\sqrt p$. 
$G$ has Shannon capacity at least $\sqrt p$, as $\{(x,ax)|x\in\mathbb F_p\}$ is independent in $G⊠G$ if $a$ is a quadratic nonresidue. 
By combining the bounds above, it follows that the Shannon capacity of $G$ is exactly $\sqrt p$.
(4) Every such $p$ is a sum of squares of two distinct natural numbers.
