Find the value of $a$ such that $F(a)=\int^{\frac \pi 2}_{0}{|\sin x-a\cos x|} \space dx$ is minimised

The logic of that first part is troubling, we shouldn't be differentiating $F$ before we've properly defined it, or with respect to a variable that isn't one of its arguments. Two ways to fix this:

We could define $G(a,\theta)=\int_0^{\theta}a\cos x-\sin x\,dx+\int_{\theta}^{\pi/2}\sin x-a\cos x\,dx$. Then $F(a)$ is the maximum of $G(a,\theta)$ with respect to $\theta\in [0,\frac{\pi}{2}]$, which we will find by differentiating $G$ with respect to $\theta$. Why the maximum? Because at each point in the interval, we're integrating either $+|\sin x-a\cos x|$ or $-|\sin x-a\cos x|$; the integral of the absolute value comes when we take the value of $\theta$ that gives us the $+$ sign everywhere.

Alternately, we solve for the splitting point $\theta$ directly. We have $\sin x-a\cos x$ when that is positive, and $a\cos x-\sin x$ when $\sin x-a\cos x$ is negative. The splitting point between them comes when $\sin x-a\cos x=0$, or $\tan x=a$.

This $\theta=\arctan a$ is always inside $[0,\frac{\pi}{2})$ for positive $a$, and the formula $F(a)=2\sqrt{a^2+1}-a-1$ we get is valid in this range. For $a<0$, that splitting point is outside the interval, and we instead get that $|\sin x-a\cos x|=\sin x + (-a\cos x)$ for all $x\in [0,\frac{\pi}{2}]$, a sum of two positive terms. Integrating this, $F(a)=1-a$ for $a<0$. So then, globally, $$F(a)=\begin{cases}2\sqrt{a^2+1}-a-1&a\ge 0\\1-a&a<0\end{cases}$$ Differentiating that, we get $F'(a)=\begin{cases}\frac{2a}{\sqrt{a^2+1}}-1&a>0\\-1&a<0\end{cases}$. That derivative is zero when $a>0$ and $\frac{2a}{\sqrt{a^2+1}}=1$, the single point $a=\frac1{\sqrt{3}}$. We must also treat $0$ where the formula changes and the endpoints $\pm\infty$ as critical points. Testing these points, we get $F(\infty)=F(-\infty)=\infty$; those certainly aren't minima. $F(0)=1$, and if we look closer $F$ is actually differentiable at zero with derivative $-1$. That isn't an extremum or even technically a critical point - but we had to look at it anyway to know. Then $$F\left(\frac1{\sqrt{3}}\right)=2\sqrt{\frac43}-\frac1{\sqrt{3}}-1=\frac{4}{\sqrt{3}}-\frac1{\sqrt{3}}-1=\sqrt{3}-1\approx 0.732$$

That value $F\left(\frac1{\sqrt{3}}\right)$ is the smallest value at any of the critical points, and is thus the minimum we sought. It looks like you made a mistake in the evaluation there that confused you. [Corrected now]

As for $-\frac1{\sqrt{3}}$? That's not a critical point, because the formula it came from doesn't apply in that region.

Suppose $a<0$. Then, over $[0,\pi/2]$, $\sin x-a\cos x\ge\sin x\ge0$, so $F(a)\ge F(0)$. Therefore $F(a)$ is not minimal.

Hence, in order to have a minimum we need $a\ge0$. In particular there exists a unique $\theta\in[0,\pi/2)$ such that $a=\tan\theta$, namely $\theta=\arctan a$. Thus we have $$ \sin x-a\cos x \begin{cases} \le 0 & 0\le x\le \arctan a \\[4px] \ge 0 & \arctan a\le x\le \pi/2 \end{cases} $$ and we can write $$ F(a)=\int_0^{\arctan a}(a\cos x-\sin x)\,dx+\int_{\arctan a}^{\pi/2}(\sin x-a\cos x)\,dx $$ Thus $$ F(a)=2a\sin\arctan a+2\cos\arctan a-a-1 $$ which is the same as you found out, but better shows that $\theta$ depends on $a$.

Computing the derivative is easy and it turns out that $$ F'(a)=\frac{2a^2}{1+a^2}\sin\arctan a+\frac{2a}{1+a^2}\cos\arctan a-1 $$ It takes just a bit of trigonometry to get $$ \sin\arctan a=\frac{a}{\sqrt{1+a^2}},\qquad \cos\arctan a=\frac{1}{\sqrt{1+a^2}} $$ and so $$ F'(a)=\frac{2a}{\sqrt{1+a^2}}-1 $$ which vanishes for $a=1/\sqrt{3}$.

By the way, this also shows that, for $a\ge0$, $$ F(a)=2\sqrt{1+a^2}-a+c $$ and, since $F(0)=1$, $c=1$.

For $a<0$, $$ F(a)=\int_{0}^{\pi/2}(\sin x-a\cos x)\,dx=1-a $$

As an alternative, consider $\theta=\arctan a$, so $a=\tan\theta$ and $$ \cos x-a\sin x=\frac{1}{\cos\theta}(\cos\theta\cos x-\sin\theta\sin x)= \frac{\cos(x+\theta)}{\cos\theta} $$ With the substitution $t=x+\theta$, your integral becomes $$ F(\tan\theta)= \frac{1}{\cos\theta}\int_{\theta}^{\theta+\pi/2}\lvert\cos t\rvert\,dt $$ The integral is simple: $$ \int_\theta^{\pi/2}\cos t\,dt-\int_{\pi/2}^{\theta+\pi/2}\cos t\,dt =2-\sin\theta-\cos\theta $$ Thus $$ G(\theta)=F(\tan\theta)=\frac{2}{\cos\theta}-\tan\theta-1 $$ and $G'(\theta)=\dfrac{2\sin^2\theta-1}{\cos^2\theta}$ vanishes for $\sin\theta=1/2$, that is, $\theta=\pi/6$. This corresponds to $a=1/\sqrt{3}$.