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If $x^2 + y^2 +\frac12 = x + y$, find $x^y+y^x$.

Now plugging our values into the expression: 

$x^y+y^x\Rightarrow \frac12^\frac12+\frac12^\frac12\\ =\sqrt{\frac12}+\sqrt{\frac12}\\ =\frac{1}{\sqrt2}+\frac{1}{\sqrt2}\\ =\frac{\sqrt2}{2}+\frac{\sqrt2}{2}=\boxed{\sqrt2}$

If $|x|=3x+1$, find $(64x^2+48x+9)^{2018}$.

Rewriting expression:

$(64x^2+48x+9)^{2018},\\ =[(8x+3)^2]^{2018},\\ =(8x+3)^{4036}.$

Solving for $x$:

$|x|=3x+1\\ x^2=(3x+1)^2\\ x^2=9x^2+6x+1\\ 8x^2+6x+1=0\\ (2x+1)(4x+1)=0\\ x_1=-0.5, x_2 = -0.25$

After plugging our values into the equation to check if they work, we will see that only $-0.25$ satisfies the equation. 

$(8\cdot(-0.25)+3)^{4036}=(-2+3)^{4036}=1^{4036}=\boxed1$