$\left(A\right)$. Given equation of parabola is $y^2+4x-2y+3=0$

By adding and subtracting one

$y^2+4x-2y+3+1-1=0$

${\left(y-1\right)}^2=-\left(4x+2\right)$

${\left(y-1\right)}^2=-4\left(x+\frac{1}{2}\right)$

Hence, vertex of parabola is $\left(-\frac{1}{2},1\right)$

$\left(B\right).$ Given, the equation of parabola, $x^2+8x+12y+4=0$

By adding and subtracting $16$

$x^2+8x+16+12y+4-16=0$

${\left(x+4\right)}^2+12\left(y-1\right)=0$

${\left(x+4\right)}^2=-12\left(y-1\right)$

Hence, vertex of parabola $\left(-4,1\right)$

$\left(C\right).$ Given, equation of parabola $y^2-x-2y+2=0$

By adding and subtracting one

$y^2-x-2y+2+1-1=0$

${\left(y-1\right)}^2-\left(x-1\right)=0$

${\left(y-1\right)}^2=\left(x-1\right)$

Here, $4a=1$

$a=\frac{1}{4}$

So, focus is $\left(1+a,1\right)$

Hence, focus is $\left(\frac{5}{4},1\right)$

$\left(D\right).$ Given the equation of parabola $x^2-2x-8y-23=0$

By adding and subtracting one

$x^2-2x+1-8y-23-1=0$

${\left(x-1\right)}^2-4\left(2y+6\right)=0$

${\left(x-1\right)}^2=-8\left(y+3\right)$

Here, $a=2$ and vertex $\left(1,-3\right)$

Hence, focus $\left(1+0,-3+2\right)$$\equiv \left(1,-1\right)$