The first and second case clearly work out, since $6789$ and $1000$ are co prime, so one can solve the given equations simultaneously.
If $x \equiv 32 \mod 1000$ then $x$ is a multiple of $8$, since $x = 1000k + 32$ so it is a sum of multiples of $8$.
However, $9876 \equiv 4 \mod 8$, so for any $x = 9876k + 54$ , it leaves a remainder of either $2$ or $6$ when divided by $8$, so it is never a multiple of $8$, so the third case does not work out.
On the other hand, the fourth case does work out, since if $x = 1000k+32 = 9876l+44$ then one may transpose and divide by four to get $250k - 2469l = 12$, and this can be solved since $250$ and $2469$ are coprime.
Hence, $1,2,4$ are solvable while $3$ is not.
A short note
More precisely, suppose you are trying to solve two simultaneous congruences, say $x \equiv a \mod b$ and $x \equiv c \mod d$. Then, write as the conclusion, $x = a+bk = c + ld$.
Therefore, rewrite this to get $ld - bk = a-c$. That is, we want to find integers $k,l$ such that $ld - bk = a-c$.
In conclusion, the question is the following: can $a-c$ be expressed as a linear combination of $b$ and $d$? For this question, Bezout's lemma gives a very clear answer : it can, if and only if $a-c$ is a multiple of the greatest common divisor of $b$ and $d$.
Therefore, all you need to do, is to check that $a-c$ is a multiple of the greatest common divisor of $b$ and $d$!
For example, in the first two parts, $(1000,6789) = 1$, and $a-c$ is a multiple of $1$ always, so the answer is yes.
In the latter two parts, we have $(1000,9876) = 4$, so it is enough to see that $a-c$ is a multiple of $4$. In the first case, $54 - 32 = 22$ isn't a multiple of $4$, while in the second case $44-32 = 12$ is.
This short note provides a complete answer to the kind of question you have been asked. It can also be extended to three or more congruences.
Best Answer
You can write $x=1000N+23$ and $x=6789M+45$ for some integers M and N. Then we seek integer solutions to
$6789M-1000N=22$.
A solution to this equation exists since we can use Euclid's algorithm to find integers $n,m $ so that
$6789m-1000n=1$,
thanks to the fact that 6789 and 1000 are comprime. Then $M=22m $ and $N=22n $. You can calculate these numbers by Euclid's algorithm.