Answer
(a) $2$
(b) $8+4\sqrt 2$
Work Step by Step
(a) Step 1. Form the geometric sequence for the side lengths: the initial side length is $a_1=1$, the next square will have a side length of $a_2=\frac{1}{\sqrt 2}=(2)^{-\frac{1}{2}}$, the third square $a_3=(2)^{-\frac{2}{2}}, ...$, and the $n$th square $a_n=(2)^{-\frac{n-1}{2}}$
Step 2. For a sequence for the area of each square ($b_n=a_n^2$): $b_1=1, b_2=(2)^{-1}, b_3=(2)^{-2}, ... b_n=(2)^{1-n}$
Step 3. The sequence $b_n$ can be identified as geometric with $r=2^{-1}$, thus the sum of the areas of all the squares is given by $S_n=\frac{1-(2^{-n})}{1-2^{-1}}$. As $n\to \infty$, $2^{-n}\to 0$, we have $S_n=2$
(b) Step 1. Form the sequence for the perimeters ($p_n=4a_n$): $p_1=4$, $p_2=4(2)^{-\frac{1}{2}}$, $p_3=4(2)^{-\frac{2}{2}}, ...$, $p_n=4(2)^{-\frac{n-1}{2}}$
Step 2. We can identify that the sequence is of geometric with $r=(2)^{-\frac{1}{2}}$, thus the sum of the perimeters of all the squares is given by: $P_n=4\times\frac{1-((2)^{-\frac{1}{2}})^n}{1-(2)^{-\frac{1}{2}}}$
As $n\to \infty$, $((2)^{-\frac{1}{2}})^n\to 0$, we have $P_n=\frac{4}{1-(2)^{-\frac{1}{2}}}=\frac{4}{1-\frac{1}{\sqrt 2}}=\frac{4\sqrt 2}{\sqrt 2-1}=\frac{4\sqrt 2(\sqrt 2+1)}{(\sqrt 2-1)(\sqrt 2+1)}=8+4\sqrt 2$
