"Assume we know that $\mathbb{Q}(2^{1/5})$ has dimension 5 as a vectorspace over $\mathbb{Q}$". What does this mean? What does it mean to consider $\mathbb{Q}(2^{1/5})$ as a vector space?
My next task is to prove that $2^{-1/5} + 1/11$ is algebraic over $\mathbb{Q}$, can i have any hints on this one?
Whenever you have a field $K$ that contains $\mathbb Q$, $K$ can always be seen as a vector space over $\mathbb Q$, the operations being the obvious ones. You can check by yourself as an exercise that all the axioms for $K$ to be a $\mathbb Q$-vector space are satisfied.
For the second part, set $x=2^{-1/5}+1/11$. Now try to compute $(x-1/11)^5$, what do you get?
Whenever we have a field extension $L/K$, then $L$ can be considered as a vector space over $K$ of dimension $[L:K]$. The axioms are obvioulsy satisfied. A basis of $L=\mathbb{Q}(2^{1/5})$ over $\mathbb{Q}$ is given by $\{1,a,a^2,a^3,a^4\}$ with $a=2^{1/5}$. So every element of $L$ is a $\mathbb{Q}$-linear combination of $1,a,a^2,a^3,a^4$.
For the second question, you may also use that the sum of two algebraic numbers is algebraic, see here. Now $2^{-1/5}$ is a root of $f(x)=2x^5-1$, and $1/11$ of $f(x)=11x-1$.
