Answer
$$\eqalign{
& \left( {\text{a}} \right){\text{Vertex: }}\left( { - 4,1} \right),{\text{Focus: }}\left( { - 7,1} \right){\text{directrix: }}x = - 1 \cr
& \left( {\text{b}} \right){\text{Vertex: }}\left( {1,\frac{1}{2}} \right),{\text{Focus: }}\left( {1,1} \right){\text{directrix: }}y = 0 \cr} $$
Work Step by Step
$$\eqalign{
& \left( {\text{a}} \right)\,\,{\left( {y - 1} \right)^2} = - 12\left( {x + 4} \right) \cr
& {\text{The equation has been written in the form }}{\left( {y - k} \right)^2} = 4p\left( {x - h} \right){\text{,}} \cr
& \underbrace {{{\left( {y - 1} \right)}^2} = - 12\left( {x + 4} \right)}_{{{\left( {y - k} \right)}^2} = 4p\left( {x - h} \right)}\,\,\, \Rightarrow \,\,\,k = 1,\,\,h = - 4,\,\,\,4p = - 12,\,\,\,p = - 3 \cr
& {\text{The vertex of a parabola is }}\left( {h,k} \right) \cr
& {\text{Vertex: }}\left( { - 4,1} \right) \cr
& {\text{The focus of a parabola is }}\left( {p + h,k} \right) \cr
& {\text{Focus: }}\left( { - 7,1} \right) \cr
& {\text{The directrix of a parabola is }}x = - p + h \cr
& {\text{directrix: }}x = - 1 \cr
& \cr
& \left( {\text{b}} \right)\,\,{\left( {x - 1} \right)^2} = 2\left( {y - \frac{1}{2}} \right) \cr
& {\text{The equation has been written in the form }}{\left( {x - h} \right)^2} = 4p\left( {y - k} \right){\text{,}} \cr
& \underbrace {{{\left( {x - 1} \right)}^2} = 2\left( {y - \frac{1}{2}} \right)}_{{{\left( {x - h} \right)}^2} = 4p\left( {y - k} \right)}\,\,\, \Rightarrow \,\,\,k = \frac{1}{2},\,\,h = 1,\,\,\,4p = 2,\,\,\,p = \frac{1}{2} \cr
& {\text{The vertex of a parabola is }}\left( {h,k} \right) \cr
& {\text{Vertex: }}\left( {1,\frac{1}{2}} \right) \cr
& {\text{The focus of a parabola is }}\left( {h,p + k} \right) \cr
& {\text{Focus: }}\left( {1,1} \right) \cr
& {\text{The directrix of a parabola is }}y = - p + k \cr
& {\text{directrix: }}y = 0 \cr
& \cr
& {\text{See the graphs below}} \cr} $$