Step 1:
Tangent and normal at $(0,1)$ to $x^2+2x-4y+4=0$
The tangent is at $(x_1,y_1)$ is
$xx_1+\large\frac{2(x+x_1)}{2}-$$4\large\frac{(y+y_1)}{2}$$+4=0$
Here $(x_1,y_1)=(0,1)$
The tangent is $0+x-2(y+1)+4=0$
(i.e) $x-2y+2=0$
Step 2:
The normal is $\perp$ to the tangent and its equation is of the form $2x+y+k=0$.It passes through $(0,1)$.
Therefore $0+1+k=0$
$\Rightarrow k=-1$
The normal is $2x+y-1=0$
Step 3:
The equation of tangent : $x-2y+2=0$
The equation of normal : $2x+y-1=0$