Consider $E = \mathbb{Q}(\sqrt{2}, \sqrt{7} )$. Here are the following questions and my respective answers follow below. I hope you have suggestions or if there are any mistakes I hope you help me fix them.

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*Find a basis for $E$ over $\mathbb{Q}(\sqrt{2}).$

*Find a basis for  $\mathbb{Q}(\sqrt{2})$ over $\mathbb{Q}.$

*Write    the general form    of  elements    of  $E.$

*Determine whether the mapping $\phi: E \rightarrow E$ such that $\phi(\sqrt{2}) = \sqrt{7}, \phi(\sqrt{7}) = \sqrt{2},$    and $\phi(q) = q, \forall q \in \mathbb{Q},$ is   an automorphism of $E$ or not.

For #1: Let $\alpha = \sqrt{7}.$ Then $\alpha^2 = 7 \Rightarrow \alpha^2 - 7 = 0.$ Thus, the degree of $\alpha$ over $\sqrt{7}$ is 2. Therefore, a basis for $E$ over $\mathbb{Q}(\sqrt{2})$ is $\{1, \sqrt{7} \}.$
For #2: $x^2 - 7$ is the irreducible polynomial of $\sqrt{2}$ in $\mathbb{Q}[x].$ Thus, a basis for $\mathbb{Q}(\sqrt{2})$ over $\mathbb{Q}$ is $\{ 1, \sqrt{2} \}.$
For #3: The general form of the elements of $E$ is $a+b\sqrt{2} + c\sqrt{7} + d\sqrt{14},$ where $a,b,c,d \in \mathbb{Q}.$
For #4:
Take the elements $1 - \sqrt{2}$ and $\sqrt{2} - \sqrt{7}$ of $\mathbb{Q}(\sqrt{2},\sqrt{7}).$ By the definition of $\phi$, $\phi(1 - \sqrt{2}) = 1 - \sqrt{7},$ and $\phi(\sqrt{2} - \sqrt{7}) = \sqrt{7} - \sqrt{2}.$ So, $\phi(1 - \sqrt{2}) \phi(\sqrt{2} - \sqrt{7}) =  \sqrt{7} - \sqrt{2} - 7 + \sqrt{2}\sqrt{7} $.
But $(1-\sqrt{2})(\sqrt{2} - \sqrt{7}) = \sqrt{2} - 2 - \sqrt{7} + \sqrt{2}\sqrt{7} \Rightarrow \phi(\sqrt{2} - 2 - \sqrt{7} + \sqrt{2}\sqrt{7} ) = \sqrt{7} - 2 - \sqrt{2} + \sqrt{7}\sqrt{2}.$
Thus, $\phi((1 - \sqrt{2})(\sqrt{2} - \sqrt{7})) \neq \phi(1 - \sqrt{2}) \phi(\sqrt{2} - \sqrt{7}).$ Therefore, $\phi$ is not an automorphism of $E = \mathbb{Q}(\sqrt{2}, \sqrt{7} )$.
Are there any comments? Are there any mistakes or can there be any improvements in my solutions?
Thank you!

1, 2, 3 look good.
Regarding 4, you can simplify by noticing that $p(x)=x^2-2 \in \mathbb Q[x]$ is a fixed polynomial under $\phi$ having $\sqrt 2$ as a root, while $\sqrt 7$ is not a root.
Or if you want to avoid using polynomials...
You have $\left(\sqrt 2\right)^2 - 2 = 0$. So applying $\phi$ you should have
$$0= \phi(0)=\phi(\left(\sqrt 2\right)^2 - 2) = \left(\phi(\sqrt 2)\right)^2 - 2 = \left(\sqrt 7\right)^2 - 2 =5$$
A contradiction implying that such automorphism can't exist.
