As $L(e^{at})=\frac1{s-a}$
So putting $a=0,L(1)=\frac1s$
and putting $a=c+id,L(e^{(c+id)t})=\frac1{s-(c+id)}$
so, $L(e^{ct}\cos dt)+iL(e^{ct}\sin dt)=\frac{s-c+id}{(s-c)^2+d^2}$
$cos^2t=\frac{1+\cos2t}2$
So, $L(cos^2t)=\frac12 L(1)+\frac12 L(\cos2t)=\frac1{2s}+\frac s{2(s^2+2^2)}$
$\cos3t\sin3t=\frac{\sin6t}2$
So, $L(\cos3t\sin3t)=\frac{L(\sin6t)}2=\frac{6}{2(s^2+6^2)}$
We know, $L(e^{at}f(t))=F(s-a)$ where $F(s)=L(f(t))$
So, $L(e^{at} t^n)=\frac{n!}{(s-a)^{n+1}}$
Hence $L(te^t)=\frac1{(s-a)^2}$ putting $a=n=1$
Putting $a=2i,n=1; L(e^{2it} t)=\frac1{(s-2i)^2}$
$L(t\cos 2t)+iL(t\sin 2t)=\frac{(s+2i)^2}{(s^2+4)^2}=\frac{s^2-4+i4s}{(s^2+4)^2}$
Now compare the real and the imaginary parts.
Observe that if $\mathcal L\{f(t)\}=F(s)$, we have $\mathcal L\left\{\mathrm e^{-ht}f(t)\right\}=F(s+h)$ and $\mathcal L\{f(kt)\}=\frac{1}{k}F\left(\frac{s}{k}\right)$ for $k>0$ and then
$$\mathcal L\left\{e^{-ht}f(kt)\right\}=\frac{1}{k}F\left(\frac{s+h}{k}\right)$$
so for $f(t)=\frac{a}{2\sqrt{\pi t^3}}\exp{(-a^2/4t)}$ with Laplcae transform $F(s)=\exp{(-a\sqrt{s})}$ we have
$$
\mathcal L\left\{e^{-ht}f(kt)\right\}=\frac{1}{k}\exp\left(-a\sqrt{\frac{s+h}{k}}\right)
$$
and then for $a=x$ we have
\begin{align}
\mathcal L^{-1}\left\{\exp\left(-x\sqrt{\frac{s+h}{k}}\right)\right\}&=e^{-ht}kf(kt)\\
&=e^{-ht}k\frac{x}{2\sqrt{\pi (kt)^3}}\exp{\left(-\frac{x^2}{4(kt)}\right)}\\
&=\frac{x}{2\sqrt{\pi kt^3}}\exp{\left(-ht-\frac{x^2}{4kt}\right)}
\end{align}
Best Answer
I am afraid there is no general relation for the laplace transform of a function.
The table you presented is quite rich in relations for well behaved families, such as $f_{a,2n}(t) = t^{2n} e^{at}$ since $\sqrt{f_{a,2n}} = f_{a/2,n}$ we get the following relation: $$ F_{\sqrt{f_{a,2n}}}(s) = F_{f_{a/2,n}}(s) = \frac{n!}{(s-a/2)^{n+1}}.$$ Moreover, if you consider functions that are constants, then you get a different relaion $F_{\sqrt{c}}(s) = \frac{\sqrt{c}}{s}$. Another different relation might be obtained if you consider a positive step functions $\phi(t) = \sum_{j=1}^k c_k 1_{[a_k,b_k)}(t)$ ($[a_k,b_k)$ disjoint ) then $\sqrt{\phi}(t)$ is again a simple function, so you can find a relation in this case too. $$F_{\sqrt{\phi}}(s) = \sum_k \sqrt{c_k}\frac{ e^{-s b_k } - e^{-sa_k}}{s}$$ All efforts seem unable to treat a general case, but you can infer from this different examples that such relation will always need some ad hoc consideration (that is, specific for each case)
Moreover, if your function is not positive, then you might have a hard time to define the square root of $f$. That is a bit discouraging
To conclude this lines, let me try to offer you a general relation that might help your efforts:
Let $f(t)>0$ for every $t$
$$ F_{\sqrt{f}}(s) = \int_0^\infty e^{-st} \sqrt{f(t)}\, dt $$
$$ F_{\sqrt{f}}(s)^2 \leq \frac{1}{s}\int_0^\infty s e^{-st} f(t)\, dt = \frac{1}{s}F_f(s)$$
where the inequality follows from Jensen's inequality ($\phi(\int f d\mu) \leq \int \phi(f) d\mu$ as long as $\phi$ is convex and $\mu$ is a positive measure with total mass $1$ and note that $\int_0^\infty s e^{-st} = 1$)