We are given $F(x)=\int_0^x f(t)\,dt$ so by the Fundamental Theorem of Calculus, $F'(x)=f(x).$

The given condition is $\int_0^{\pi} (f'(x)+F(x))\cos x\,dx=2.$

Split the integral: $\int_0^{\pi} f'(x)\cos x\,dx + \int_0^{\pi} F(x)\cos x\,dx =2.$

Evaluate the first integral by integration by parts: $\int_0^{\pi} f'(x)\cos x\,dx = [f(x)\cos x]_0^{\pi} - \int_0^{\pi} f(x)(-\sin x)\,dx$ $= [f(x)\cos x]_0^{\pi} + \int_0^{\pi} f(x)\sin x\,dx.$

Now, $[f(x)\cos x]_0^{\pi}=f(\pi)\cos\pi - f(0)\cos 0 = (-6)(-1)-f(0)(1)=6-f(0).$

Hence, $\int_0^{\pi} f'(x)\cos x\,dx = 6-f(0)+\int_0^{\pi} f(x)\sin x\,dx.$

Evaluate the second integral by integration by parts: $\int_0^{\pi} F(x)\cos x\,dx = [F(x)\sin x]_0^{\pi} - \int_0^{\pi} F'(x)\sin x\,dx.$

Since $\sin 0=\sin \pi=0$, the boundary term is zero. Also $F'(x)=f(x)$. Therefore, $\int_0^{\pi} F(x)\cos x\,dx = -\int_0^{\pi} f(x)\sin x\,dx.$

Add the two results: $\int_0^{\pi} (f'(x)+F(x))\cos x\,dx$ $=\left(6-f(0)+\int_0^{\pi} f(x)\sin x\,dx\right)+\left(-\int_0^{\pi} f(x)\sin x\,dx\right)$ $=6-f(0).$

Given this integral equals $2$, we get $6-f(0)=2.$ Therefore, $f(0)=4.$

Comparison with stored answer: Stored correct answer = $4$, which matches our derived result.