Can $\frac{a^4}{b^3}$ be arbitary close to $1$?
Regarding dot number 2 and 3: Find such integers $a$ and $b$ so $a^4=b^3$. It's rather obvious that there are infinitely many of them. Just set $a=c^3$ and $b=c^4$ for some integer $c$. Then the ratio $\frac{(a+1)^4}{b^3}$ is arbitrarily close to $1$. Analogously, $\frac{(a-1)^4}{b^3}$ is close from below. This also gives a method of generating examples.
Edit regarding dot number one:
If $n=1$ I have found a proof that no such solution exist. $$ a^4 - b^3 = 1 \Leftrightarrow b^3 = a^4 - 1 = (a^2 + 1)(a^2 - 1) = (a^2 + 1)(a+1)(a-1) $$ I will show that a common divisor of any two of three factors on the right side is at most $2$. Indeed, by Euclid's algorithm $$gcd(a-1, a+1) = gcd(a-1, 2) \in \{ 1, 2\}$$ $$gcd(a-1, a^2+1) = gcd(a-1, a^2 + 1 - 2(a-1)) =$$$$= gcd(a-1, (a-1)^2+2) = gcd(a-1, 2) \in \{ 1, 2\}$$ The third proof is analogous.
Moreover, if $a$ is even, all three of these equal $1$ while for odd $a$, all three equal $2$. If $a$ is odd, then consider $b' = \frac b 2$. Else just set $b' = b$. This yields that integral $b'^3$ is a product of coprime integers, thus all three of them are a cube. However $a+1$ and $a-1$ differ only by $2$, while $\frac{a+1}2$ and $\frac{a-1}2$ differ by $1$, therefore they can't be both cubes, since the least difference between consecutive cubes is $2^3 - 1^3 = 7$.
On question 2.
Taking $$a=k^3-3,\\b=k^4-4k,$$ for $k\in\mathbb{N}$, $k>4$, one can construct sequence of pairs $(a,b)$, for which $$a^4 = b^3+n,$$ where $$n = 6k^6-44k^3+81,$$ so $$n=6a^2-8a+3<6a^2.$$
Therefore, one can construct infinitely many pairs $(a,b)$ which provide error estimation $$\left|\dfrac{b^3}{a^4}-1\right|<\dfrac{6}{a^2}.$$
Other sequence of such kind:
$$a=(3k)^3-1,\\b=(3k)^4-4k,$$ provides better error estimation, but with the same asymptotic (~ $a^{-2}$).
Consider $a,b$ coprime positive numbers. There are $2$ cases:
$$a^4+d=b^3,\tag{1}$$ $$b^3+d=a^4.\tag{2}$$
According to ABC conjecture, there are finitely many pairs $(a,b)$ such that $$d<{a^{5/3-\epsilon}},\tag{3}$$ where $\epsilon>0$.
Explanation.
Denote $C = \max\{a^4,b^3\}$.
$(1)$ and $(2)$ are expressions of the form $$A+B=C.$$
Refer now to to ABC conjecture.
Denote $$R = \mathrm{rad}(ABC),$$ where $\mathrm{rad}(\bullet)$ is radical of an integer.
Then $$R = \mathrm{rad}(abd) \le abd,$$ $$\log R \le \log a + \log b + \log d.$$
Denote $x = C^{1/12}$. Denote $y = \log x = \dfrac{1}{12}\log C$. Then $\log a \approx 3y$, $\log b \approx 4y$.
Consider value $q$ ("quality"):
$$q = \dfrac{\log C}{\log R} \ge \dfrac{\log C}{\log a+ \log b + \log d}\approx \dfrac{12 y}{3y+4y+\log d}.$$
If condition $(3)$ is true, then $\log d < (5-3\epsilon)y$, and $q$ is greater than $1$ (with some gap).
Your example has $\log d \approx 4.7785 y$, which provides quality $1.0188$ for ABC-triple $(9825757^4,137318688623, 2104527924^3)$.
This implies that for any $n\ne 0$ there are finitely many coprime pairs$ (a,b)$ s.t. $a^4-b^3=n$. (if ABC conjecture is true, of course).