Step 1: Analyze option (1).
The radius of a nucleus is given by
\[
R=R_0A^{1/3}
\]
The volume of the nucleus is therefore
\[
V=\frac{4}{3}\pi R^3
\]
Substituting \(R=R_0A^{1/3}\),
\[
V\propto (A^{1/3})^3
\]
\[
V\propto A
\]
The mass of the nucleus is also proportional to \(A\).
Hence,
\[
\text{Density}=\frac{\text{Mass}}{\text{Volume}}
\]
Since both mass and volume are proportional to \(A\),
\[
\text{Density}=\text{constant}
\]
Therefore, nuclear density is approximately independent of mass number \(A\).
Thus, option (1) is correct.
Step 2: Analyze option (2).
The radius of a nucleus is not directly proportional to \(A\).
Instead,
\[
R\propto A^{1/3}
\]
Hence, option (2) is incorrect.
Step 3: Analyze option (3).
Binding energy is directly proportional to mass defect according to Einstein’s relation:
\[
E=\Delta mc^2
\]
Thus, larger mass defect means larger binding energy.
Therefore, binding energy is not inversely proportional to mass defect.
Hence, option (3) is incorrect.
Step 4: Analyze option (4).
When heavy nuclei split into lighter nuclei during nuclear fission, energy is released, not absorbed.
This happens because the binding energy per nucleon increases for medium mass nuclei.
Hence, option (4) is incorrect.
Step 5: Final conclusion.
Therefore, the correct statement is
\[
\boxed{\text{The nuclear density in general is independent of mass number }A}
\]