How do we solve an equation of the form $a+x=b$ in the real numbers? We add $-a$ to both sides to get $x=b-a$. What does "$-a$" mean? $-a$ is the unique real number that, when added to $a$, will give you $0$.
So in essence, what we are doing is: $\begin{align*} a+x &= b &\text{(assumption)}\\ -a+(a+x) &= -a+b&\\ (-a+a)+x &= -a+b &(+\text{ is associative)}\\ 0+x &= -a+b&\text{(definition of }-a\text{)}\\ x & = -a+b &\text{(property of }0\text{)} \end{align*}$
We do the same thing in (i) and (ii). Only... what is "$-A$"? It's the unique subset of $U$ such that $A+A=\varnothing$.
So... what is the element which, when added to $A$, gives $\varnothing$? Once you find it, you can solve both $A+X=0$ and $A+X=B$, in terms of $A$, $B$, and $0$.
For (iii), you need to remember that $A-B$ means $A+(-B)$; so again, the first step is to figure out what is $-B$ in terms of $B$. Once you do, you can compute $A-B$, and compute $A+B$ (using the definition of $+$), and then compute their product (aka their intersection). This will give you $X$ in terms of $A$ and $B$.
Added. Henning raises a good point in comments: that in (iii), the $-$ on the right hand side might refer not to the $-$ operation in the ring, but rather to the set-theoretic difference of $A$ and $B$. This would be bad use of potentially confused and confusing notation; if that's what it means, then you would first compute $A\setminus B = \{a\in A\mid a\notin B\}$, then compute the "product" (intersection) with $A+B$ (aka their symmetric difference) and obtain an expression for $X$ from that.