Write the first integral: substitute $u=t-x$, $du=dt$. Then ${\int }_0^x\tan (t-x)dt={\int }_{-x}^0\tan udu=-{\int }_0^{-x}\tan udu$. Equivalently, ${\int }_0^x\tan (t-x)dt=-{\int }_0^x\tan tdt$ via direct rewrite.

So $f(x)=-{\int }_0^x\tan tdt-{\int }_0^{x}f(t)\tan tdt$.

Differentiate (Leibniz): $f^{\prime }(x)=-\tan x-f(x)\tan x=-\tan x(1+f(x))$.

Let $y=f(x)+1$. Then $y^{\prime }=-y\tan x$, so $\frac{dy}{y}=-\tan xdx$. Integrating: $\ln |y|=\ln |\cos x|+C\Rightarrow y=K\cos x$.

Initial: $f(0)=0$ (both integrals vanish at $x=0$), so $y(0)=1\Rightarrow K=1$. Therefore $f(x)=\cos x-1$.

Then $f^{\prime }(x)=-\sin x$ and $f^{\prime \prime }(x)=-\cos x$.

$f^{\prime \prime }(\pi /6)=-\cos (\pi /6)=-\frac{\sqrt{3}}{2}$.

$f^{\prime }(-\pi /6)=-\sin (-\pi /6)=\frac{1}{2}$.

$f(\pi /6)=\cos (\pi /6)-1=\frac{\sqrt{3}}{2}-1$.

Sum: $-\frac{\sqrt{3}}{2}+12\cdot \frac{1}{2}+\frac{\sqrt{3}}{2}-1=6-1=5$.