Step 1:
Equation of the tangent to $2xy-2x-8y-1=0$ at $(x_1,y_1)$
$\large\frac{1}{2}$$\times 2(xy_1+yx_1)-2\large\frac{(x+x_1)}{2}$$-8\large\frac{(y+y_1)}{2}$$-1=0$
Here $(x_1,y_1)=(-2,\large\frac{1}{4})$
The tangent is $(\large\frac{x}{4}-$$2y)-(x-2)-4(y+\large\frac{1}{4})$$-1=0$
$x-8y-4x+8-16y-4-4=0$
$-3x-24y=0$
$3x+24y=0$
$3(x+8y)=0$
$\Rightarrow x+8y=0$
Step 2:
The normal is $\perp$ to the tangent.Its equation is of the form $8x-y+k=0$
It passes through $(-2,\large\frac{1}{4})$.
Therefore $8(-2)-\large\frac{1}{4}$$+k=0$
$k=16+\large\frac{1}{4}=\large\frac{65}{4}$
The equation of the normal is $8x-y+\large\frac{15}{4}$$=0$
$\Rightarrow 32x-4y+65=0$
Step 3:
Equation of tangent $x+8y=0$
Equation of normal $32x-4y+65=0$