1. Convert the complex line into Cartesian form

Let $z=x+iy, \qquad \bar z=x-iy.$

Given $z(1+i)+\bar z(1-i)=4.$

Now, $z(1+i)=(x+iy)(1+i)=x+ix+iy+i^2y=(x-y)+i(x+y),$

and $\bar z(1-i)=(x-iy)(1-i)=x-ix-iy+(-i)(-iy)=(x-y)-i(x+y).$

Adding, $z(1+i)+\bar z(1-i)=2(x-y).$

So the curve is $2(x-y)=4 \implies x-y=2.$

Thus the given curve is the straight line $x-y=2.$

2. Interpret the region $|z-3|\le 1$

Since $3$ means the complex number $3+0i$, the region $|z-3|\le 1$ is the circle with center $(3,0)$ and radius $1$.

So we need the areas into which the line $x-y=2$ divides this disk.

3. Find the distance of the center from the line

The line is $x-y-2=0.$

Distance of point $(3,0)$ from this line is $d=\frac{|3-0-2|}{\sqrt{1^2+(-1)^2}}=\frac{1}{\sqrt2}.$

Since $d=\frac{1}{\sqrt2}<1,$ the line cuts the circle into two unequal parts.

4. Use the formula for area cut by a chord

For a circle of radius $r$, if a chord is at distance $d$ from the center, then the area of the smaller segment is $A_s=r^2\cos^{-1}\!\left(\frac{d}{r}\right)-d\sqrt{r^2-d^2}.$

Here, $r=1,\qquad d=\frac{1}{\sqrt2}.$

Hence $A_s=\cos^{-1}\!\left(\frac{1}{\sqrt2}\right)-\frac{1}{\sqrt2}\sqrt{1-\frac12}.$

Now, $\cos^{-1}\!\left(\frac{1}{\sqrt2}\right)=\frac{\pi}{4},$

and $\sqrt{1-\frac12}=\frac{1}{\sqrt2}.$

Therefore,