For each part, show that the real number $\alpha$ defined in that part is algebraic over $\mathbb{Q}$.
Squaring both sides and subtracting $2$, we get $\alpha^2 -2 = \sqrt{3 + \sqrt{5}}$. Squaring both sides again and subtracting $3$, we get $(\alpha^2 - 2)^2 - 3 = \sqrt{5}$. Finally, squaring both sides and subtracting $5$, we get $((\alpha^2-2)^2-3)^2 - 5 = 0$, so $\alpha$ is a zero of a nonzero polynomial with rational coefficients, namely $((x^2-2)^2 - 3)^2 - 5$.
Squaring both sides, we get $\alpha^2 = 2 + 2\sqrt{2}\sqrt{3} + 3 = 5 + 2\sqrt{6}$. Subtracting $5$ and squaring again, we get $(\alpha^2-5)^2 = 4 \cdot 6 = 24$. Therefore $\alpha$ is a zero of the polynomial $(x^2-5)^2 - 24$.
Note: we don't have to expand these polynomials, because we can already see that they are nonzero and have rational coefficients.
Let $F$ be a field, let $n$ be a positive integer, and let $p(x) \in F[x]$ be an irreducible polynomial of degree $n$. Define $E = F[x]/\langle p(x) \rangle$, which is an extension field of $F$. Define the element $\alpha \in E$ by $\alpha = x + \langle p(x) \rangle$.
Let $\beta \in E$, say $\beta = f(x) + \langle p(x) \rangle$ where $f(x) \in F[x]$. By the division algorithm for polynomials over a field (Theorem 23.1) there are unique polynomials $q(x),r(x) \in F[x]$ such that $\deg r(x) \lt n$ and $f(x) = p(x)q(x) + r(x)$.
In fact we will only need uniqueness for $r(x)$, not $q(x)$: there is a unique polynomial $r(x) \in F[x]$ such that $\deg r(x) \lt n$ and $f(x) = p(x)q(x) + r(x)$ for some $q(x) \in F[x]$. Note that the statement "$f(x) = p(x)q(x) + r(x)$ for some $q(x) \in F[x]$" is equivalent to the statement $f(x) + \langle p(x) \rangle = r(x) + \langle p(x) \rangle$. Therefore there is a unique polynomial $r(x) \in F[x]$ such that $\deg r(x) \lt n$ and $f(x) + \langle p(x) \rangle = r(x) + \langle p(x) \rangle$.
Recall that $f(x) + \langle p(x) \rangle = \beta$ and $r(x) + \langle p(x) \rangle = r(x + \langle p(x) \rangle) = r(\alpha)$. So we have shown that there is a unique polynomial $r(x) \in F[x]$ such that $\deg r(x) \lt n$ and $\beta = r(\alpha)$.
A polynomial is determined by its sequence of coefficients, and a polynomial of degree less than $n$ is determined by a sequence of length $n$, namely the sequence $a_0,\ldots,a_{n-1}$ of coefficients of $x^0,\ldots,x^{n-1}$. In particular, the number of polynomials in $\mathbb{Z}_p[x]$ of degree less than $n$ is equal to the number of sequences of length $n$ of elements in $\mathbb{Z}_p$, which is $p^n$. By part a, this means that $E$ has order $p^n$ in the case $F = \mathbb{Z}_p$.
Define $p(x) = x^4 + x + 1 \in \mathbb{Z}_2[x]$ and define the factor ring $E = \mathbb{Z}_2[x]/\langle p(x) \rangle$. Define the element $\alpha = x + \langle p(x) \rangle \in E$.
Note that $p(0) = 1$ and $p(1) = 1$, so $p(x)$ has no zeroes in $\mathbb{Z}_2$. Therefore $p(x)$ cannot be factored as the product of a degree $1$ polynomial and a degree $3$ polynomial in $\mathbb{Z}_2[x]$. It remains to show that $p(x)$ cannot be factored as the product of two degree $2$ polynomials in $\mathbb{Z}_2[x]$. Suppose toward a contradiction that $p(x) = (ax^2 + bx + c)(dx^2 + ex + f)$ where $a,b,c,d,e,f \in \mathbb{Z}_2$. Then $ad$ is the coefficient of $x^4$ in $p(x)$, which is $1$, and $cf$ is the coefficient of $x^0$ in $p(x)$, which is $1$, so $a,d,c,f = 1$ and $$p(x) = (x^2 + bx + 1)(x^2 + ex + 1) = x^4 + (b+e)x^3 + (be+2)x^2 + (b+e)x + 1.$$ This contradicts the fact that the coefficients of $x^3$ and $x$ in $p(x)$ are different.
$E = \{0,1,x,x+1,x^2,x^2+1,x^2+x,x^2+x+1,x^3,x^3+1,x^3+x,x^3+x+1,x^3+x^2,x^3+x^2 + 1,x^3+x^2 + x,x^3+x^2+x+1\}$.
We have $\alpha^4 + \alpha + 1$, so $\alpha^4 = -\alpha - 1 = \alpha + 1$. So the powers of $\alpha$ are $\alpha^0$, $\alpha^1$, $\alpha^2$, $\alpha^3$, $\alpha^4 = \alpha + 1$, $\alpha^5 = (\alpha+1)\alpha = \alpha^2 + \alpha$, and we can stop here because by Lagrange's theorem the order of $\alpha$ in the group $E^*$ divides the order of $E^*$, which is $15$, and we have shown that the order of $\alpha$ is not $1$, $3$, or $5$. (You can also just compute all the powers up to $\alpha^{14}$.)
Define $\beta = 2^{1/3} + 2^{2/3} = \alpha + \alpha^2$ where $\alpha = 2^{1/3}$. Using the fact that $\alpha^3 = 2$, we can compute the first few powers of $\beta$ as follows:
Then we see that $\beta^3 - 6\beta - 6 = 0$, so $\beta$ is a zero of the polynomial $p(x) = x^3 - 6x - 6$. We claim that $p(x)$ is irreducible in $\mathbb{Q}[x]$, which will imply that $\operatorname{irr}(\beta,\mathbb{Q}) = p(x)$ and $\operatorname{deg}(\beta,\mathbb{Q}) = \deg p(x) = 3$. Because $p(x)$ has degree $\le 3$, it suffices to show that $p(x)$ has no zeroes in $\mathbb{Q}$. Suppose toward a contradiction that $p(x)$ has a rational zero $a/b$, where we assume without loss of generality that $\gcd(a,b) = 1$.
Because $p(x)$ has integer coefficients, we must have $a \mid 6$ and $b \mid 1$ by the rational root theorem, so $a/b$ is $\pm 1$, $\pm 2$, $\pm 3$, or $\pm 6$. It's easy to check that none of these values is a zero of $p(x)$, especially if we rewrite the equation $p(x) = 0$ as $x^3 = 6(x+1)$. If we evaluate at $\pm 1$, $\pm 2$, or $\pm 3$ then the left hand side is not divisible by $6$, and if we evaluate at $\pm 6$ then the left hand side is much too large in absolute value.
Note that $F[\alpha] = \phi_\alpha[F[x]]$ where $\phi_\alpha : F[x] \to E$ is the evaluation homomorphism at $\alpha$. The range of a ring homomorphism is a subring of the codomain, so $F[\alpha]$ is a subring of $E$. It remains to prove that $F[\alpha]$ is a field.
By the fundamental homomorphism theorem, we have $F[\alpha] \cong F[x]/N$ where $N = \ker(\phi_\alpha)$. Note that the ideal $N$ is nontrivial because $\alpha$ is algebraic. Note also that the ideal $N$ is prime: for all $f(x),g(x) \in F[x]$, if $f(x)g(x) \in N$ then $f(\alpha)g(\alpha) = 0$, so $f(\alpha) = 0$ or $g(\alpha) = 0$, so $f(x) \in N$ or $g(x) \in N$. Another way to see this is that $\{0\}$ is a prime ideal of $E$, and the inverse image of a prime ideal under a homomorphism is a prime ideal (or else the improper ideal, but that doesn't happen in this case.) Both ways use the fact that $E$ is an integral domain.
Because $F[x]$ is a principal ideal domain, and every nontrivial prime ideal of a principal ideal domain is maximal, it follows that $N$ is a maximal ideal of $F[x]$. Therefore $F[x]/N$ is a field, and so is $F[\alpha]$.