Inspired by "calculator's worst enemy"

$10^{32}$ , or $(10^{16})^{2}$ is the smallest 33-digit perfect square; therefore, the biggest 32-digit perfect square would be $(10^{16}-1)^{2}$ , or 99999999999999980000000000000001 (there are fifteen "9"s and "0"s). See this if you wonder how we get this number.

Now, the sum of its digits is $9 \times 15 + (8+1)$ , or $9 \times 16 =144$ . Hence, $a = \boxed {144}$ . Since the digit "0" occurs in the number, the product of the number's digit is $\boxed {0}$ .

Here comes the final step: plug in 144 and 0 respectively for $a$ and $b$ . We get

$a^{b} + b^{a} = 144^{0} + 0^{144} =1+0= \boxed {1}$

see rules of exponents if you need help on calculating with exponents!