If $\text{Gal}(K/\mathbb{Q})\cong Z_5$, then show $K(\sqrt{2})/\mathbb{Q}$ is Galois

a) Find a field extension $$K/\mathbb{Q}$$ such that $$\text{Gal}(K/\mathbb{Q})\cong Z_5$$ ($$Z_5$$ denotes the cyclic group on $$5$$ elements).

b) Let $$L=K(\sqrt{2})$$. Show $$L/\mathbb{Q}$$ is Galois and determine the cardinality of $$\text{Gal}(L/\mathbb{Q})$$.

c) Give the isomorphism type of $$\text{Gal}(L/\mathbb{Q})$$ explicitly and describe the automorphisms explicitly.

My thoughts: For part (a), we already know $$\mathbb{Q}(\zeta_{11})/\mathbb{Q}$$ is Galois with $$\text{Gal}(\mathbb{Q}(\zeta_{11})/\mathbb{Q})\cong Z_{10}$$. Also, $$Z_{10}$$ has $$Z_2$$ as its unique normal subgroup of index $$5$$, and thus $$Z_2$$ corresponds to a Galois extension $$K/\mathbb{Q}$$ with $$\text{Gal}(K/\mathbb{Q})\cong Z_{10}/Z_2\cong Z_5$$.

For part (b), I'm not sure if I can just use the fact that $$L$$ is a quadratic extension of $$K$$, or if I need to explicitly determine $$K$$ first and find a polynomial for which $$L=K(\sqrt{2})$$ is its splitting field. I think I can determine $$K$$ explicitly (as in finding a suitable generator to adjoin to $$\mathbb{Q}$$) using my work from part (a), but this might be working too hard.

For part (c), I think this will follow quickly once I have part (b), but I am again unsure.

So $$K$$ is the splitting field of a separable polynomial $$f(x)$$.
You know that $$K(\sqrt{2})$$ is the splitting field of the polynomial $$f(x)(x^2-2)$$; this is a separable polynomial so $$K(\sqrt{2})$$ is Galois.
You know that $$\sqrt{2}$$ can't be in $$K$$ because the degree of $$K$$ on $$\mathbb{Q}$$ is 5, so you know that the degree of the intersection of $$K$$ and $$\mathbb{Q}[\sqrt{2}]$$ is 1 and the degree of $$K$$ is $$\frac{5\cdot 2}{1}=10$$.
For part (c) you know that that the Galois group of $$L$$ contains two different normal subgroup, one related to $$\mathbb{Q}[\sqrt{2}]$$ and the other related to $$K$$. So you have that $$Gal(L/\mathbb{Q})$$ is a group of cardinality 10 with two different normal subgroup of cardinality 2 and 5 with trivial intersection. You can say that it's isomorphic to $$Z_{10}$$.