A reflection formula for the Hurwitz zeta function, which does not seem to be well known, uses half of the polynomials generated by $\frac{1}{-1+\sqrt{t-1}\cot(\sqrt{t-1}u)}$. (Look at the sections "formula" and "example"). For instance, the coefficient of $\frac{u^6}{6!}$ is $p_6(t)=32t+416t^2+272t^3$ and we have indeed $$\zeta(7,x)-\zeta(7,1-x)=\frac{(\pi\cot\pi x)^{7}}{6!}\left(\frac{32}{\cos^2\pi x}+\frac{416}{\cos^4\pi x}+\frac{272}{\cos^6\pi x}\right),$$ generally $$\boxed{\zeta(2n+1,x)-\zeta(2n+1,1-x)=\frac{(\pi\cot\pi x)^{2n+1}}{(2n)!}p_{2n}(\sec^2\pi x)}.$$

Note that the first argument of $\zeta(\cdot,\cdot)$ is an odd integer (and interestingly the formula involves odd powers of $\pi$). I wondered if this formula still works when it is even, i.e. when $n$ is a half-integer. And the funny answer is: almost... Taking e.g. $n=5/2$ and $x=1/10$, the corresponding formula (using $p_5(t)=16t+88t^2+16t^3$) yields 1950.3076... instead of the LHS of 1950.30017..., which does not seem to have a (known) closed form.

Is there a way to explain this numerical closeness?

Looking at the reflection identity from a different angle (no pun intended):
Define rationals $a_n,b_n,c_n$ by $$\frac1{1+2\cos\sqrt{t}}=:\sum_{n=0}^\infty a_nt^n,\ \frac1{-1+2\cos\sqrt{t}}=:\sum_{k=0}^\infty b_nt^n,\ \frac1{2\cos\sqrt{t}}=:\sum_{k=0}^\infty c_nt^n$$ (for $a_n$, see A279120 and A279121, for $b_n$, also called Glaisher's H' numbers, see A002114. while the $c_n$ are, up to a factor $ 2\cdot(2n)!$, the (absolute) Euler numbers). Then we have the particular instances

$$\zeta(2n+1,\tfrac13)-\zeta(2n+1,\tfrac23)=\frac{\sqrt{3}}2(2\pi)^{2n+1}a_n$$ $$\zeta(2n+1,\tfrac16)-\zeta(2n+1,\tfrac56)=\frac{\sqrt{3}}2(2\pi)^{2n+1}b_n$$ $$\zeta(2n+1,\tfrac14)-\zeta(2n+1,\tfrac34)= (2\pi)^{2n+1}c_n.$$

In this form, nothing relates to half-integers $n$, but as a special case, expressions of the form $\zeta(2n,\tfrac13)-\zeta(2n,\tfrac23)$ occur in this question, also bearing a $\sqrt{3}$ factor. (One could say that the ones with even argument, having a closed form, are "resolved" into the rest there). May this help in understanding what is going on here?

A reflection formula for the Hurwitz zeta function, which does not seem to be well known, uses half of the polynomials generated by $\frac{1}{-1+\sqrt{t-1}\cot(\sqrt{t-1}u)}$. (Look at the sections "formula" and "example"). For instance, the coefficient of $\frac{u^6}{6!}$ is $p_6(t)=32t+416t^2+272t^3$ and we have indeed $$\zeta(7,x)-\zeta(7,1-x)=\frac{(\pi\cot\pi x)^{7}}{6!}\left(\frac{32}{\cos^2\pi x}+\frac{416}{\cos^4\pi x}+\frac{272}{\cos^6\pi x}\right),$$ generally $$\boxed{\zeta(2n+1,x)-\zeta(2n+1,1-x)=\frac{(\pi\cot\pi x)^{2n+1}}{(2n)!}p_{2n}(\sec^2\pi x)}.$$

Note that the first argument of $\zeta(\cdot,\cdot)$ is an odd integer (and interestingly the formula involves odd powers of $\pi$). I wondered if this formula still works when it is even, i.e. when $n$ is a half-integer. And the funny answer is: almost... Taking e.g. $n=5/2$ and $x=1/10$, the corresponding formula (using $p_5(t)=16t+88t^2+16t^3$) yields 1950.3076... instead of the LHS of 1950.30017..., which does not seem to have a (known) closed form.

Is there a way to explain this numerical closeness?

Looking at the reflection identity from a different angle (no pun intended):
Define rationals $a_n,b_n,c_n$ by $$\frac1{1+2\cos\sqrt{t}}=:\sum_{n=0}^\infty a_nt^n,\ \frac1{-1+2\cos\sqrt{t}}=:\sum_{k=0}^\infty b_nt^n,\ \frac1{2\cos\sqrt{t}}=:\sum_{k=0}^\infty c_nt^n$$ (for $a_n$, see A279120 and A279121, for $b_n$, also called Glaisher's H' numbers, see A002114. while the $c_n$ are, up to a factor $ 2\cdot(2n)!$, the (absolute) Euler numbers). Then we have the particular instances

$$\zeta(2n+1,\tfrac13)-\zeta(2n+1,\tfrac23)=\frac{\sqrt{3}}2(2\pi)^{2n+1}a_n$$ $$\zeta(2n+1,\tfrac16)-\zeta(2n+1,\tfrac56)=\frac{\sqrt{3}}2(2\pi)^{2n+1}b_n$$ $$\zeta(2n+1,\tfrac14)-\zeta(2n+1,\tfrac34)= (2\pi)^{2n+1}c_n.$$

In this form, nothing relates to half-integers $n$, but as a special case, expressions of the form $\zeta(2n,\tfrac13)-\zeta(2n,\tfrac23)$ occur in this question, also bearing a $\sqrt{3}$ factor. (One could say that the ones with even argument, having a closed form, are "resolved" into the rest there). May this help in understanding what is going on here?