To find the relationship between $\mathrm{K}_1$ and $\mathrm{K}_2$, let's carefully analyze the given equilibrium reactions and their constants.

First, let's write down the equilibrium reactions clearly:

Reaction 1: $$3 \mathrm{~A}_2+\mathrm{~B}_2 \rightleftharpoons 2 \mathrm{~A}_3 \mathrm{~B}$$ with equilibrium constant $\mathrm{K}_1$.

Reaction 2: $$\mathrm{~A}_3 \mathrm{~B} \rightleftharpoons \frac{3}{2} \mathrm{~A}_2 + \frac{1}{2} \mathrm{~B}_2$$ with equilibrium constant $\mathrm{K}_2$.

Now, understand that $\mathrm{K}_2$ represents the equilibrium constant of the reverse reaction of Reaction 1 but with both sides divided by 2. We need to relate these constants. Here is the step-by-step process:

We know Reaction 1 is: $$3 \mathrm{~A}_2 + \mathrm{~B}_2 \rightleftharpoons 2 \mathrm{~A}_3 \mathrm{~B}$$.

The equilibrium constant for Reaction 1 is: $$\mathrm{K}_1 = \frac{[\mathrm{A}_3 \mathrm{~B}]^2}{[\mathrm{A}_2]^3[\mathrm{B}_2]}$$.

For Reaction 2: $$\mathrm{~A}_3 \mathrm{~B} \rightleftharpoons \frac{3}{2} \mathrm{~A}_2 + \frac{1}{2} \mathrm{~B}_2$$, the equilibrium constant $\mathrm{K}_2$ can be expressed as the equilibrium constant of the reverse of Reaction 1, with adjusted coefficients.

The equilibrium constant for the reverse reaction of Reaction 1 would be $$\frac{1}{\mathrm{K}_1}$$. Since Reaction 2 includes halving the coefficients, the equilibrium constant should be adjusted by taking the square root:

$$\mathrm{K}_2 = \left(\frac{1}{\mathrm{K}_1}\right)^{1/2} = \frac{1}{\sqrt{\mathrm{K}_1}}$$.

Therefore, the relationship between $\mathrm{K}_1$ and $\mathrm{K}_2$ is:

Option D: $$\mathrm{K}_2 = \frac{1}{\sqrt{\mathrm{K}_1}}$$.