解：$(1)$由$ρ=\frac {m}{V}$得，溢出水的体积$V_{水}=\frac {m_{水}}{ρ_{水}}=\frac {300\ \mathrm {g}}{1.0\ \mathrm {g/cm}^3}=300\ \mathrm {cm}^3；$该摆件的体积等于溢出水的体积，即$V_{摆件}=V_{水}=300\ \mathrm {cm}^3$
$(2)$若该摆件是实心的，由$ρ=\frac {m}{V}$得，该摆件的质量$m=ρ_{铜}V_{摆件}=8.9\ \mathrm {g/cm}^3×300\ \mathrm {cm}^3=2670\ \mathrm {g}$
$(3)$由$ρ=\frac {m}{V}$得，摆件中铜的体积$V_{实心}=\frac {m'}{ρ_{铜}}=\frac {890\ \mathrm {g}}{8.9\ \mathrm {g/cm}^3}=100\ \mathrm {cm}^3；$空心部分体积$V_{空}=V_{摆件}-V_{实心}=300\ \mathrm {cm}^3−100\ \mathrm {cm}^3=200\ \mathrm {cm}^3$
