For the "harder" (??) second part of the question, let $a=i$ or let $a=\frac{-1+i\sqrt{3}}{2}$.
Added: The first part of the question is (for me at least) more difficult than the second part. Maybe I am missing something obvious. The solution below uses some algebra, but not Galois Theory, just degrees of extensions.
We prove something that looks stronger but isn't. Let $a$ be a real number. If there exists a positive integer $n$, and a non-zero rational $e$, such that $a^n$ and $(e+a)^n$ are rational, then $a$ is rational. Suppose the result is not correct. Then there is a smallest positive integer $n$, a real irrational $a$, and a non-zero rational $e$ such that $a^n$ and $(e+a)^n$ are rational. It is clear that $n$ must be $\ge 2$.
First we do something completely unnecessary. By assumption $a^n$ and $(e+a)^n$ are rational. Bring these rationals to a common denominator, which can be taken to be a perfect $n$-th power $r^n$. If $n$ is even, then $a^n=\frac{p}{r^n}$ and $(e+a)^n=\frac{q}{r^n}$ for some non-negative integers $p$ and $q$. If $n$ is odd, then, depending on the signs of $a$ and $e+a$, $a^n=\pm\frac{p}{r^n}$ and $(e+a)^n=\pm\frac{q}{r^n}$ for some non-negative integers $p$ and $q$. Then in the even case, $(ar)^n=p$ and $(r+ar)^n=q$, and in the odd case we have the same thing, with $p$ and/or $q$ possibly decorated with minus signs.
Let $w=ar$. Then $w=p^{1/n}$ and $r+w=q^{1/n}$ in the even case, and $w=\pm p^{1/n}$, $r+w=\pm q^{1/n}$ in the odd case. Note that $w$ is a real irrational. Note also the crucial fact that by the minimality of $n$, there is no positive integer $m such that simultaneously $(p^{1/n})^m$ and $(q^{1/n})^m$ are rational.
Since $\pm p^{1/n}$ and $\pm q^{1/n}$ differ by an integer $r$, they have the same degree. By the minimality of $n$, this degree is $n$. But $p^{1/n}=q^{1/n}+r$. Take the $n$-th power of both sides. We find that $q^{1/n}$ is the root of a polynomial with integer coefficients, of degree $, contradicting the fact that $q^{1/n}$ has degree $n$.
Comment: My first posted "proof" implicitly assumed that $a>0$. Thanks to Matt E for pointing out that modification was needed.