Every element $x \in A$ is $x\leq2$(why?) Thus $2$ is an upper bound of $A$. Can you find a lower upper bound of $A$? The smallest of the upper bounds is the $\sup A$.
Every element $x \in A$ is $x\geq-1$(why?) Thus $-1$ is a lower bound of $A$. Can you find a larger lower bound of $A$? The largest of the lower bounds is the $\inf A$.
Edit: You got it right $\sup A=2$ and $\inf A=-1.$
To prove this you must show that $2$ is an upper bound of $A$ (i.e. $x\leq 2, \ \forall x \in A$ ) and that for any $\epsilon>0, \ 2-\epsilon$ is not an upper bound of $A$(i.e. there is an $x \in A$ with $x>2-\epsilon$) . SImilar with $\inf A$.
First of all, let me note that the terms minimum, maximum, infimum and supremum do not live in a vacuum. They are properties that a certain element can have with respect to a given order. Now forget about this and let us only focus on the order $(\Bbb R; \le)$ that you seem to be interested in. Given any subset of real numbers $S \subseteq \Bbb R$, we may ask whether or not there is some $r \in \Bbb R$ such that $r$ is bigger then all the elements in $S$. If such an $r$ exists, we call it an upper bound for $S$.
For example: Let $S = (0,1)$. Every element $s \in S$ is smaller than $5$ and hence $5$ is an upper bound for $S$. However, there are smaller upper bounds. $4$ is an upper bound for $S$ as well and so is $\pi$ and $\sqrt{2}$. We may hence ask if there is a least upper bound for $S$ and in fact, we have that $1$ is an upper bound for $S$ and no $r < 1$ is an upper bound for $S$. Hence $1$ is the least upper bound for $S$, which we also call the supremum of $S$.
We designed the real numbers in such a way that once a given subset of reals $S \subseteq \Bbb R$ has any upper bound, it will always have a (unique!) least upper bound, i.e. a supremum. (A supremum is, by definition, the same as a least upper bound.) Let me stress that this is by construction of the reals, it is not true for other orders. For example, in $(\Bbb Q; \le)$, the set $\{ q \in \Bbb Q \mid q^2 < 2\}$ clearly has an upper bound, but it doesn't have a supremum.
Let us return to $(\Bbb R; \le)$. Note that $\{ x \in \Bbb R \mid 0 < x \}$ does not have an upper bound and thus does not have a supremum. Recall that any subset $S \subseteq \Bbb R$ has a supremum if and only if it has an upper bound.
So what about maxima? A maximum of a set of reals $S \subseteq \Bbb R$ is a least upper bound of $S$ that is also an element of $S$. Since suprema are unique (provided they exist), we thus have that a maxima of $S$ is also the supremum of $S$ and that $S$ has at most one maximum.
Let's have a second look at $S = (0,1)$. We already know that $S$ has $1$ as it's suprema, but $1 \not \in (0,1)$. This immediately implies that $S$ does not have a maximum. (By the argument above, if $S$ had a maximum, it would be equal to its supremum. But the supremum of $S$ is not an element of $S$ and hence not a maximum.)
I'll leave the case for infima of $S \subseteq \Bbb R$ (which are greatest lower bounds of $S$) and minima of $S$ (which are greatest lower bounds of $S$ that are also elements of $S$) to you.
Best Answer
Consider the sequence $$a_n=1-\sum_{m=1}^n\frac{1}{4^m}$$ Note that $E=\{a_n\}_n$, where that sequence is strictly decreasing and bounded. Therefore you have a maximum (because its strictly decreasing) and an infimum with no minimum because if it ever reached the minimum (suppose it's reached by $a_{n_0}$), it contradicts the fact it is the minimum because the sequence is strictly decreasing.