The task is to determine $Gal(L|K)$ for $L=\mathbb{F}_2(X)$ and $K=\mathbb{F}_2(X^2)$
I'm not sure what $K$ looks like. I assume $\mathbb{F}_2(X^2)\cong \{ \frac{P(X)}{Q(X)}: P,Q \in F_2[X^2] \}$.
My first attempt was to guess a $K$-basis of $L$, namely $(1,X)$ but it didn't work because I can't get odd exponents in the denominator. Therefore I assume it is an extension of degree $>2$.
On the other hand I also realized that $irr(X,\mathbb{F}_2(X^2))=T^2-X^2$, hence $[\mathbb{F}_2(X^2)[X]:\mathbb{F}_2(X^2)]=2$ but here I get confused.
If $\mathbb{F}_2(X^2)[X]$ is a field, then it should be closed under inverse and thus it contains $\frac{1}{X}$. Does this mean $\mathbb{F}_2(X^2)[X]=\mathbb{F}_2(X)$?
I'd be very thankful if you could clarify the situation.
You're right that $(1,X)$ is a $K$-basis of $L$.
First notice that $\mathbb{F}_2[X]\subset L$. For $P\in\mathbb{F}_2[X]$, one can write $$P=\sum_{}p_{2k}X^{2k} + \left(\sum_{}p_{2k+1}X^{2k}\right)X$$
Then remember that the Frobenius is $\mathbb{F}_2$-linear : this implies, among other things, that for $P\in\mathbb{F}_2[X]$, $P(X^2)=P(X)^2$.
The result follows : for $(P,Q)\in\mathbb{F}_2[X]^2$ and $Q\neq 0$, just write $$\frac{P(X)}{Q(X)}=\frac{P(X)Q(X)}{Q(X)^2}=\frac{P(X)Q(X)}{Q(X^2)}$$
which is therefore in the $K$ vector space spanned by $(1,X)$ inside $L$.