Solution:

By definition, a pair of positive integers $(a, b)$ is nice if there exist positive integers $k, h$ such that $b=k a$ and $b+1=h(a+1)$. In particular, we have $k a+1=h(a+1)$ and therefore
$$h=\frac{k a+1}{a+1} .$$
Since $h$ is an integer, we have that $k a+1=k(a+1)-(k-1)$ is divisible by $a+1$, from which $k-1$ is divisible by $a+1$ and therefore there exists a nonnegative integer $n$ such that $k-1=n(a+1)$. Putting everything together, we obtain that if the pair $(a, b)$ is nice then there exists an integer $n$ such that $b=k a=a(n a+n+1)$.
On the other hand, if $b=a(n a+n+1)$ for some nonnegative integer $n$, then, clearly, $a$ divides $b$ and $b+1=(n a+n+1) a+1=n(a+1) a+(a+1)$ is divisible by $a+1$.
Summing up, the pair $(a, b)$ is nice if and only if $b=a(n a+n+1)$ for some positive integer $n$.

a. Given $a$, there exist infinitely many positive integers $b$ for which the pair $(a, b)$ is nice: indeed, it follows from what was said above that for every positive integer $n$ the pair $(a, a(n a+n+1))$ is nice.

Second solution: for every odd positive integer $d$, the pair of the type $(a, a^{d})$ is nice: clearly $a$ divides $a^{d}$; moreover $a^{d}+1=(a+1)(1-a+a^{2}-a^{3}+\ldots+a^{d-1})$ is divisible by $a+1$. We have thus obtained another infinite family of nice pairs $(a, b)$ with $a$ given.

b. Note that, if $m>n$, then $a(m a+m+1)>a(n a+n+1)$. Since the nice pairs are all and only those of the type $(a, a(n a+n+1))$ (for $n$ a nonnegative integer) and since for $n=0$ we have $a(n a+n+1)=a$, the minimum integer $b>a$ for which $(a, b)$ is nice is obtained for $n=1$, that is $b=a^{2}+2 a>a$.

Second solution: Given $a$, the minimum $b$ (such that $b>a$ and $(a, b)$ is a nice pair) is a multiple of $a$ of the type $m a$ ($m>1$ integer). I observe that if $m a+1$ gives remainder $r$ upon division by $a+1$, then $(m+1) a+1=(m a+1)+(a+1)-1$ gives remainder $r-1$. Since for $m=1$ the remainder of the division of $m a+1$ by $a+1$ is zero, the next multiple of $a$ to have this property will be given when $m=a+2$, that is $b=(a+2) a$.

c. From what was said above, we know that the pair $(18, b)$ is nice if and only if $b=18(18 n+n+1)=18(19 n+1)$. The condition
$$20=a+2 \mid b+2=342 n+20,$$
is satisfied if and only if 20 divides $342 n$ or, equivalently, if 10 divides $n$. We thus deduce that the minimum $b=18(19 n+1)$ greater than 18 for which the pair $(18, b)$ is nice and $b+2$ divides $a+2=20$ is obtained for $n=10$, that is $b=191 \cdot 18=3438$.