For each part, compute the sum and product of the given polynomials in the given polynomial ring.
$f(x) + g(x) = 2 + 5x$ and $f(x)g(x) = 1 + 5x$.
$f(x) + g(x) = 1 + 6x + 3x^2 + x^3$ and $f(x)g(x) = 4x + x^2 + 6x^3 + 2x^4 + 3x^5$.
Let $R$ be a ring and let $f(x),g(x) \in R[x]$.
First consider the case where $f(x)$ and $g(x)$ are nonzero. Let $a_mx^m$ and $b_nx^n$ be the leading terms of $f(x)$ and $g(x)$ respectively. Then $a_mb_nx^{m+n}$ is the leading term of $f(x)g(x)$. Note that the coefficient $a_mb_n$ is nonzero because $R$ has no zero divisors. Therefore $\deg f(x)g(x) = m+n = \deg f(x) + \deg g(x)$.
Next consider the case where at least one of $f(x)$ and $g(x)$ is zero. Without loss of generality we may assume that $f(x) = 0$. Then $f(x)g(x) = 0$ and we have $\deg f(x)g(x) = -\infty = -\infty + \deg g(x) = \deg f(x) + \deg g(x)$.
In problem 1a, $\deg f(x)g(x) = 1 \ne 1 + 1 = \deg f(x) + \deg g(x)$ (but $\mathbb{Z}_6$ is not an integral domain; in particular the leading coefficients $2$ and $3$ are zero divisors.)
Assume that $R[x]$ has a unity $\sum_{i = 0}^n a_ix^i$. We claim that $a_0$ is a unity for $R$. Let $a \in R$. Then $ax^0 \in R[x]$. (Typically we denote $ax^0$ simply by $a$, but let's not do that now.) We have $ax^0 = ax^0 \cdot \sum_{i = 0}^n a_ix^i = \sum_{i = 0}^n (a\cdot a_i)x^i$. In particular the coefficients of $x^0$ on both sides are equal: $a = a \cdot a_0$. A similar argument shows that $a_0 \cdot a = a$, so $a$ is a unity for $R$.
Let $p$ be a prime. Then $\mathbb{Z}_p$ is an integral domain, which implies that the polynomial ring $\mathbb{Z}_p[x]$ is an integral domain. Note that $\mathbb{Z}_p[x]$ has the same characteristic as $\mathbb{Z}_p$ (namely $p$) and is infinite because it contains the distinct elements $1,x,x^2,\ldots.$ It is not a field, but because it is an integral domain it has a field of quotients, which we denote by $\mathbb{Z}_p(x)$. Note that the field of quotients $\mathbb{Z}_p(x)$ of $\mathbb{Z}_p[x]$ has the same characteristic as $\mathbb{Z}_p[x]$ (namely $p$) and is also infinite because there is an injection from $\mathbb{Z}_p[x]$ to $\mathbb{Z}_p(x)$ given by $f(x) \mapsto f(x)/1$.
Assume that the kernel of $\phi_\alpha$ contains a polynomial of degree $1$, say $f(x) = a_0 + a_1x$ where $a_1 \ne 0$. Then $\phi_\alpha(f(x)) = a_0 + a_1\alpha = 0$, so $a_1\alpha = -a_0$, and because $F$ is a field we can write $\alpha = -a_0 a_1^{-1} \in F$.
Conversely, assume that $\alpha \in F$. Define $f(x) = x- \alpha$. Then $\deg f(x) = 1$ and $\phi_\alpha(f(x)) = \alpha - \alpha = 0$, so $f(x) \in \ker(\phi_\alpha)$.
One example of a nontrivial automorphism of $\mathbb{Z}[x]$ is the evaluation homomorphism $\phi_{-x} : \mathbb{Z}[x] \to \mathbb{Z}[x]$. We have $$\phi_{-x} \left( \sum_{i=0}^n a_ix^i \right ) = \sum_{i=0}^n a_i(-x)^i = a_0 - a_1x + a_2x^2 - a_3x^3 + \cdots.$$ That is, $\phi_{-x}$ negates the coefficients of odd powers of $x$ and does not change the coefficients of even powers of $x$. Because the function $\phi_{-x}$ has an inverse (namely itself) it is bijective, and we already know that it is a homomorphism from $\mathbb{Z}[x]$ to $\mathbb{Z}[x]$. Therefore it is an automorphism of $\mathbb{Z}[x]$. It is not the identity function because $\phi_{-x}(x) = -x \ne x$.
(There are other examples also.)