This question is pretty old, but based on its number of views, it probably deserves a more robust answer. In order to show that this limit exists, we must show that the left-handed limit is equal to the right-handed limit. But before we do that, let's make the following observation:
$$\frac{\sqrt{|x|}}{x} = \begin{cases} \frac{1}{\sqrt{x}}, \quad &\text{if } x>0\\ -\frac{1}{\sqrt{-x}}, \quad &\text{if } x<0\end{cases}$$
This observation arises from the fact that the numerator is always positive, but the denominator is the same sign as $x$.
With this in mind, we calculate the left and right-handed limits.
$$\text{LEFT:} \lim_{x \to 0^-}\frac{\sqrt{|x|}-\sqrt{|0|}}{x-0} = \lim_{x \to 0^-} \frac{\sqrt{|x|}}{x} = \lim_{x \to 0^-} -\frac{1}{\sqrt{-x}}=-\frac{1}{\sqrt{\text{small pos. number}}} = -\infty$$
$$\text{RIGHT:} \lim_{x \to 0^+}\frac{\sqrt{|x|}-\sqrt{|0|}}{x-0} = \lim_{x \to 0^+} \frac{\sqrt{|x|}}{x} = \lim_{x \to 0^+} \frac{1}{\sqrt{x}}=\frac{1}{\sqrt{\text{small pos. number}}} = +\infty$$
Since the left-handed limit and the right-handed limit are not the same, the limit does not exist, and therefore, the function is not differentiable at $x=0$.
The function defined by
$$f(x)=\begin{cases}\frac{\sin(x)}{x}&,x\ne 0\\\\1&,x=0\end{cases}$$
is not only differentiable at $x=0$, it is continuously differentiable there.
NOTE:
I thought it would be instructive to present a way forward that relies only on a standard, elementary inequality and the squeeze theorem. To that end, we proceed.
The derivative at $x=0$ is given by
$$f'(0)\equiv \lim_{h\to 0}\frac{\frac{\sin(h)}{h}-1}{h} \tag 1$$
Recalling from elementary geometry that the sine function satisfies the inequalities
$$\cos(h) \le \frac{\sin(h)}{h}\le 1 \tag 2$$
for $|h|\le \pi/2$, we see that the term under the limit in $(1)$ satisfies the inequalities
$$-2\sin^2(h/2)= \cos(h)-1\le \frac{\sin(h)}{h}-1\le 0 \tag 3$$
Then, taking absolute values, dividing by $|h|$, and using the right-hand side inequality in $(2)$ yields
$$0 \le \left|\frac{\frac{\sin(h)}{h}-1}{h}\right|\le \frac12 |h| \tag 4$$
whereupon applying the squeeze theorem to $(4)$ produces the limit
$$\lim_{h\to 0}\frac{\frac{\sin(h)}{h}-1}{h}=0$$
Therefore, $f'(0)=0$.
For $x\ne 0$, we have
$$f'(x)=\frac{x\cos(x)-\sin(x)}{x^2}$$
To see that $f'(x)$ is continuous at $x=0$, we need to show that
$$\lim_{x\to 0}\frac{x\cos(x)-\sin(x)}{x^2}=0 $$
Again, using $(2)$ we see that
$$0\le \left|\frac{x\cos(x)-\sin(x)}{x^2}\right|\le\left|\frac{1-\cos(x)}{x}\right| \le \frac12 |x| \tag 5$$
whereupon applying the squeeze theorem to $(5)$ produces the limit
$$\lim_{x\to 0}\frac{x\cos(x)-\sin(x)}{x^2}=0 $$
Therefore,
$$\lim_{x\to 0}f'(x)=0=f'(0)$$
which shows that $f'$ is continuously differentiable at $0$.
Best Answer
In order to be differentiable everywhere, $f$ must be continuous. As $f$ is a piecewise continuous function:
To solve for $a$, in the last step using the fact that $a$ is positive $$\lim_{x\rightarrow0^-}f(x)=\lim_{x\rightarrow0^+}f(x)$$ $$\lim_{x\rightarrow0^-}(a^2+e^x)=\lim_{x\rightarrow0^+}(2+x)$$ $$a^2+e^0=2+0$$ $$a^2=1$$ $$a=1$$
Similarly, we can solve for $c$ in terms of $b$: $$\lim_{x\rightarrow3^-}f(x)=\lim_{x\rightarrow3^+}f(x)$$ $$\lim_{x\rightarrow3^-}(2+x)=\lim_{x\rightarrow3^+}(c-\frac{b^2}{x})$$ $$2+3=c-\frac{b^2}{3}$$ $$c=5+\frac{b^2}{3}$$
Now, we take into account that $f$ is differentiable at $x=3$ (and using the fact that $b$ is positive): $$\lim_{x\rightarrow3^-}f'(x)=\lim_{x\rightarrow3^+}f'(x)$$ $$\lim_{x\rightarrow3^-}1=\lim_{x\rightarrow3^+}\frac{b^2}{x^2}$$ $$1=\frac{b^2}{9}$$ $$b=3$$ $$c=5+\frac{b^2}{3}=5+3=8$$
Therefore $\frac{a+b+c}{2}=\frac{1+3+8}{2}=6$