Answer
See explanations.
Work Step by Step
(a) Step 1. Rewrite the function as $f(x)=\frac{1}{2}f(x)+\frac{1}{2}f(x)+\frac{1}{2}f(-x)-\frac{1}{2}f(-x)=\frac{1}{2}(f(x)+f(-x))+\frac{1}{2}(f(x)-f(-x))$
Step 2. Let $E(x)=\frac{1}{2}(f(x)+f(-x))$ and $O(x)=\frac{1}{2}(f(x)-f(-x))$
Step 3. Check if $E(x)$ is an even function: $E(-x)=\frac{1}{2}(f(-x)+f(x))=E(x)$, thus $E(x)$ is even.
Step 4. Check if $O(x)$ is an odd function: $O(-x)=\frac{1}{2}(f(-x)-f(x))=-O(x)$, thus $O(x)$ is odd.
Step 5. Conclusion: function $f$ can be expressed as the sum of an even function and an odd function.
(b) Step 1. Assume there is a different way of writing $f$ as the sum of an even and an odd function, and let $f(x)=E_1(x)+O_1(x)$ where $E_1$ is even and $O_1$ is odd.
Step 2. With the above assumption, we have $E+O=E_1+O_1$ or $E-E_1=O_1-O$.
Step 3. Let $g(x)=E-E_1=O_1-O$, as the difference of two even (odd) functions will result in another even (odd) function, thus we have $g(x)$ being both even and odd.
Step 4. Based on the result from Exercise 11, we know that $g(x)=0$, thus we have $E=E_1$ and $O=O_1$ which means that there is only one way to write $f$ as the sum of an even and an odd function.
