To evaluate the contour integrals, you can simply parameterize. For the first example, set $z=w+r e^{i \phi}$, $\phi \in [-\pi,\pi]$. Then the integral is equal to
$$r^{1/2} \int_{-\pi}^{\pi} d\phi \, e^{i \phi/2} \left (w-a+r e^{i \phi}\right ) = 4 (w-a) r^{1/2} - \frac{4}{3} r^{3/2} $$
The other integral may be done similarly.
It would be the situation in B: you would deform around the pole. It works as follows.
The inverse Laplace transform is given by Cauchy's theorem. I present the parametrization of each piece of the contour, assuming that the radius of the semicircular detour about the pole $z=-1$ and the branch point $z=0$ is $\epsilon$:
$$\int_{c-i \infty}^{c+i \infty} ds \frac{e^{s t}}{\sqrt{s} (1+s)} + e^{i \pi} \int_{\infty}^{1+\epsilon} dx \frac{e^{-t x}}{e^{i \pi/2} \sqrt{x} (1-x)}+i \epsilon \int_{\pi}^0 d\phi \, e^{i \phi} \frac{e^{t (-1+ \epsilon e^{i \phi})}}{\sqrt{e^{i \pi}+\epsilon e^{i \phi}} (\epsilon e^{i \phi})}\\+ e^{i \pi} \int_{1-\epsilon}^{\epsilon} dx \frac{e^{-t x}}{e^{i \pi/2} \sqrt{x} (1-x)}+i \epsilon \int_{\pi}^{-\pi} d\phi \, e^{i \phi} \frac{e^{t \epsilon e^{i \phi}}}{\sqrt{\epsilon e^{i \phi}} (1+\epsilon e^{i \phi})} +e^{-i \pi} \int_{\epsilon}^{1-\epsilon} dx \frac{e^{-t x}}{e^{-i \pi/2} \sqrt{x} (1-x)}\\+ i \epsilon \int_{2 \pi}^{\pi} d\phi \, e^{i \phi} \frac{e^{t (-1+ \epsilon e^{i \phi})}}{\sqrt{e^{-i \pi}+\epsilon e^{i \phi}} (\epsilon e^{i \phi})}+ e^{-i \pi} \int_{1+\epsilon}^{\infty} dx \frac{e^{-t x}}{e^{-i \pi/2} \sqrt{x} (1-x)} = 0$$
Note that the integrals about the semicircular detours above and below the axis (the 3rd and the 7th integrals) cancel. In the limit as $\epsilon \to 0$, the integral about the branch point (the 5th integral) also vanishes. We are then left with, as $\epsilon \to 0$,
$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \frac{e^{s t}}{\sqrt{s} (1+s)} + \frac1{2 \pi} PV \int_{\infty}^0 dx \frac{e^{-t x}}{\sqrt{x} (1-x)} - \frac1{2 \pi} PV \int_0^{\infty} dx \frac{e^{-t x}}{\sqrt{x} (1-x)} = 0$$
where $PV$ denotes the Cauchy principal value of the integral. Thus, the ILT is given by (subbing $x=u^2$)
$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \frac{e^{s t}}{\sqrt{s} (1+s)} = \frac1{\pi} PV \int_{-\infty}^{\infty} du \, \frac{e^{-t u^2}}{1-u^2} $$
To evaluate the integral, we rewrite as
$$e^{-t} PV \int_{-\infty}^{\infty} du \, \frac{e^{t (1- u^2)}}{1-u^2} = e^{-t} I(t)$$
where
$$I'(t) = e^{t} PV \int_{-\infty}^{\infty} du \, e^{-t u^2} = \sqrt{\pi} t^{-1/2} e^{t} $$
and $I(0) = 0$. Thus,
$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \frac{e^{s t}}{\sqrt{s} (1+s)} = e^{-t}\frac1{\pi} \sqrt{\pi} \int_0^t dt' \, t'^{-1/2} e^{t'} = e^{-t} \frac{2}{\sqrt{\pi}} \int_0^{\sqrt{t}} dv \, e^{v^2} $$
or, finally,
$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \frac{e^{s t}}{\sqrt{s} (1+s)} = e^{-t} \operatorname{erfi}{\left (\sqrt{t} \right )} $$
Best Answer
Indeed, choose the branch of $\sqrt{1-z^2}$ so that its branch cut is $[-1,1]$. For example, we may let
$$\sqrt{1-z^2} = \exp\left(\frac{1}{2}\log(z-1)+\frac{1}{2}\log(z+1) - \frac{i\pi}{2}\right),$$
where $\log(\cdot)$ is the principal complex logarithm. Then for $-1 < x < 1$,
\begin{align*} \lim_{\epsilon \to 0^+} \sqrt{1-(x+i\epsilon)^2} &= \sqrt{1-x^2} \\ \lim_{\epsilon \to 0^+} \sqrt{1-(x-i\epsilon)^2} &= -\sqrt{1-x^2}, \end{align*}
So, the integral of $ f(z) = \frac{1}{(z+2)\sqrt{1-z^2}} $ along the dog bone contour (red contour in the image below)
is $2I$:
$$ \oint_{\color{red}{\text{dogbone}}} f(z) \, \mathrm{d}z = 2I. $$
So, if $R > 2$ and $\epsilon > 0$ is small enough, then
\begin{align*} \oint_{|z| = R} f(z) \, \mathrm{d}z &= \oint_{|z+2| = \epsilon} f(z) \, \mathrm{d}z + \oint_{\color{red}{-\text{dogbone}}} f(z) \, \mathrm{d}z \\ &= 2\pi i \, \underset{z=-2}{\mathrm{Res}} \, f(z) - \oint_{\color{red}{\text{dogbone}}} f(z) \, \mathrm{d}z \\ &= \frac{2\pi}{\sqrt{3}} - 2I \end{align*}
As you see, you first forgot to take account of the pole of $f(z)$ at $z=-2$. When its contribution is properly accounted as in the computation above, then by using that
$$ \oint_{|z| = R} f(z) \, \mathrm{d}z = \lim_{R\to\infty} \oint_{|z| = R} f(z) \, \mathrm{d}z = 0, $$
we end up with
$$ I = \frac{\pi}{\sqrt{3}}. $$