Closed form for a zeta series :$\sum^\infty_{k=2}\frac{(-1)^{k-1}\zeta(k)}{(k+2)2^{k+2}}$

As mentioned by Claude Leibovici, you have

$$ \sum^\infty_{k=2}\frac{(-1)^{k-1}\zeta(k)}{(k+2)2^{k+2}}=-\frac{1}{8}-\frac{\gamma}{24}+\frac{\ln 2}{12}+\frac{\ln A}{2}-\frac{7\zeta(3)}{16 \pi ^2}. \tag1 $$

Here is a hint.

From the classic identity verified by the digamma function $\displaystyle \psi:=\Gamma'/\Gamma$, wich may be obtained from the Euler product giving $\Gamma(x+1)$, you have $$ \psi(x+1) = -\gamma + \sum_{k=1}^{\infty}\frac{x}{k(k+x)}\quad x\neq 0,-1,-2,-3,\dots$$ then you easily obtain, for $|x|<1$, $$ \begin{align} \psi(x+1) & = -\gamma + \sum_{k=1}^{\infty}\frac{x}{k^2}\frac{1}{1+\dfrac{x}{k}} \\ &= -\gamma + \sum_{k=1}^{\infty}\frac{1}{k^2}\sum_{n=0}^{\infty}\frac{(-1)^n}{k^n}x^{n+1} \\ &= -\gamma - \sum_{n=0}^{\infty}(-1)^{n-1} \zeta(n+2){x^{n+1}} \\ &= -\gamma - \sum_{k=2}^{\infty}(-1)^{k-1} \zeta(k){x^{k-1}} \\ \end{align} $$ and $$-\gamma x^2 - x^2\psi(x+1) = \sum_{k=2}^{\infty}(-1)^{k-1} \zeta(k){x^{k+1}}. \tag2$$

Using $(2)$ gives

$$ \begin{align} \sum^\infty_{k=2}\frac{(-1)^{k-1}\zeta(k)}{(k+2)2^{k+2}}&=\sum_{k=2}^{\infty}(-1)^{k-1} \zeta(k)\int_0^{1/2}\!\!x^{k+1}dx\\ &= \int_0^{1/2}\!\sum_{k=2}^{\infty}(-1)^{k-1} \zeta(k){x^{k+1}}\:dx \\ &= -\gamma \int_0^{1/2}\!x^2 dx - \int_0^{1/2}\! x^2\psi(x+1)\:dx \\ &=-\frac{\gamma}{24} - \int_0^{1/2}\! x^2\psi(x+1)\:dx, \end{align} $$ then integrating by parts twice leads to $$ \begin{align} \sum^\infty_{k=2}\frac{(-1)^{k-1}\zeta(k)}{(k+2)2^{k+2}} & = -\frac{\gamma}{24} - \frac14\log \Gamma\left(\frac32\right)+2\int_0^{1/2}\! x\log \Gamma(x+1)\:dx\\ &= -\frac{\gamma}{24} - \frac18\log \pi- \frac14\ln2+\zeta'\left(-1,\frac32\right)+2\int_0^{1/2}\! \zeta'(-1,x+1)\:dx \\ &= -\frac{\gamma}{24}-\frac{1}{8}+\frac{\ln 2}{12}+\frac{\ln A}{2}-\frac{7\zeta(3)}{16 \pi ^2}, \end{align} $$ where we have used the identity (25.11.34) and special values of $\zeta'(s,a)$.