Work with the total height sum $S$ and the class size $n$ directly, instead of jumping straight to the average.
The average height is $A = S/n$. Statement I says removing the tallest student (height $56$) drops the average by $1$:
\[\frac{S-56}{n-1} = \frac{S}{n} - 1\]Cross-multiplying and simplifying (the $S/n$ terms cancel out the same way whichever order you substitute) reduces this to a single relation between $n$ and $A = S/n$, namely $n+A=57$. This is one equation with two unknowns, so $n$ and $A$ individually stay undetermined from Statement I alone.
Statement II says removing the shortest student (height $42$) raises the average by $1$:
\[\frac{S-42}{n-1} = \frac{S}{n} + 1\]which reduces in the same way to $n-A=41$, again one equation with the same two unknowns, so Statement II alone is equally insufficient on its own.
But the two conditions together give two independent equations in the two unknowns $n$ and $A$: $n+A=57$ and $n-A=41$. Two independent linear equations in two unknowns have exactly one solution, found by adding them: $2n=98$, so $n=49$ students, and then $A = 57-49 = 8$ cm.
Since each statement alone leaves an underdetermined system, but together they form a solvable 2-by-2 linear system with a unique answer, the two statements together are necessary and sufficient, while neither is enough alone. This matches option (3): the average height, \(\boxed{8 \text{ cm}}\), needs both statements combined.