What are all the intermediate fields of $\mathbb{Q}\big(\sqrt{3+\sqrt{5}}\big)$ containing $\mathbb{Q}$?

First of all all your work seems correct to me (there are some typos where you defined the automorphisms)

The intermediate fields:

We note first that $ord(\sigma)=ord(\tau)=ord(\sigma\tau)=2$, hence their index is also $2$.

1) For $\sigma$:

$\sigma(\alpha_1)=-\alpha_1$, hence the square of $\alpha_1$ lies in the fixed field of $\sigma$. We have $\alpha_1^2=3+\sqrt{5}$, hence $\mathbb Q(3+\sqrt{5})=\mathbb Q(\sqrt{5})$ is the fixed field of $\sigma$ because, $\sqrt{5}$ has degree $2$ over $\mathbb Q$.

2) For $\tau$:

$\tau(\alpha_1)=\alpha_3$, hence the sum of both lies in the fixed field of $\tau$: $\alpha_1+\alpha_3=\sqrt{3+\sqrt{5}}+\sqrt{3-\sqrt{5}}=\sqrt{a}$. Squaring both sided yields that $a=3+\sqrt{5}+2\sqrt{4}+3-\sqrt{5}=6+2\sqrt{4}=10$. So the fixed field of $\tau$ is $\mathbb Q(\sqrt{10}$), because $\sqrt{10}$ has degree $2$ over $\mathbb Q$.

3) For $\sigma\tau$:

$\sigma\tau(\alpha_1)=\alpha_4$, hence the sum of both lies in the fixed field of $\sigma\tau:$ $\alpha_1+\alpha_4=\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}=\sqrt{b}$. Squaring both sided again yields: $3+\sqrt{5}-2\sqrt{4}+3-\sqrt{5}=6-4=2$. So the fixed field of $\sigma\tau$ is $\mathbb Q(\sqrt{2})$, because $\sqrt{2}$ has degree 2 over $\mathbb Q$.

I hope there are no mistakes, typos etc. (I am on the way) but thats a way you can solve some of this problems.