Prove that opposite sides of parallelogram PQRS are equal and calculate x and angle E for parallelogram EFGH given angles G and F.

Given that PQRS is a parallelogram, we know that $PQ \parallel RS$ and $PR \parallel QS$.
Draw a diagonal PS.
Consider $\triangle PQS$ and $\triangle RSP$:

• $\angle QPS = \angle RSP$ (Alternate interior angles, since $PQ \parallel RS$)
• $\angle QSP = \angle RPS$ (Alternate interior angles, since $PR \parallel QS$)
• $PS = SP$ (Common side) Therefore, $\triangle PQS \cong \triangle RSP$ (Angle-Side-Angle congruence criterion). Since the triangles are congruent, their corresponding sides are equal:
• $PQ = RS$
• $QS = PR$ Thus, the opposite sides of the parallelogram are equal.

In a parallelogram, consecutive angles are supplementary. Therefore, $\hat{G} + \hat{F} = 180^\circ$.
Substitute the given expressions for $\hat{G}$ and $\hat{F}$:

$$(9x + 12^\circ) + (15x - 24^\circ) = 180^\circ$$ $$24x - 12^\circ = 180^\circ$$ $$24x = 180^\circ + 12^\circ$$ $$24x = 192^\circ$$ $$x = \frac{192}{24}$$ $$x = 8$$

The value of $x$ is $\boxed{\text{8}}$.

In a parallelogram, opposite angles are equal. Therefore, $\hat{E} = \hat{G}$.
First, calculate the measure of $\hat{G}$ using the value of $x$ found in Step 2:

$$\hat{G} = 9x + 12^\circ$$ $$\hat{G} = 9(8) + 12^\circ$$ $$\hat{G} = 72^\circ + 12^\circ$$ $$\hat{G} = 84^\circ$$

Since $\hat{E} = \hat{G}$:

$$\hat{E} = 84^\circ$$

The size of $\hat{E}$ is $\boxed{\text{84}^\circ}$.