Laplace transform of $ t^{1/2}$ and $ t^{-1/2}$

For $t^{-1/2}$ we have

$$F(s)=\int\limits_0^\infty e^{-st} t^{-1/2}dt$$ Now make $st = u$ so that

$$F(s)=s^{-1/2} \int\limits_0^\infty e^{-u} u^{-1/2}du$$

Since the integral is $\Gamma(1/2)$ we get

$$F(s)=s^{-1/2} \sqrt \pi=\sqrt{\frac{\pi}{s}}$$

Why don't you want to prove the general case? Use the best tools you have when you can. We have

$$\mathcal{L}(t^n)=\int\limits_0^\infty e^{-st}t^n dt$$

We make $st = u$ and get

$$\mathcal{L}(t^n)=\frac{1}{s^{n+1}}\int\limits_0^\infty e^{-u}u^n du$$

Thus

$$\mathcal{L}(t^n)=\frac{\Gamma(n+1)}{s^{n+1}}$$

Like I mentioned earlier, there is the rule $\mathcal{L}\{tf(t)\}=-F'(s)$, here applicable with $f(t)=t^{-1/2}$.

Or just directly apply $d/ds$ to part (a). Integration by-parts is equivalent ($u=t^{1/2},dv=e^{-ts}dt$).