解：?$(1) ②$?已知?$f(2)=5，$??$f(2n)=25$?
∵?$f(2n)=f(2 + 2+···+2)(n$?個?$2$?相加?$)$?
根據(jù)?$f(\mathrm {m})·f(\mathrm {n})=f(m + n)，$?可得?$f(2n)=f(2)× f(2)×··· ×f(2)=5^{n}$?
又∵?$f(2n)=25 = 5^2$?
∴?$n = 2$?
?$ (2) $?已知?$f(\mathrm {a})=3，$?化簡?$f(\mathrm {a})·f(2a)·f(3a)···f(10a)$?
?$ f(2a)=f(a + a)=f(\mathrm {a})·f(\mathrm {a})=f^2(\mathrm {a})$?
?$ f(3a)=f(2a + a)=f(2a)·f(\mathrm {a})=f^2(\mathrm {a})·f(\mathrm {a})=f^3(\mathrm {a})$?
?$ f(10a)=f^{10}(\mathrm {a})$?
∴?$f(\mathrm {a})·f(2a)·f(3a)···f(10a)=f(\mathrm {a})·f^2(\mathrm {a})·f^3(\mathrm {a})···f^{10}(\mathrm {a})$?
根據(jù)同底數(shù)冪相乘，底數(shù)不變，指數(shù)相加，?$1 + 2+···+10=\frac {10×(1 + 10)}2=55$?
則?$f(\mathrm {a})·f^2(\mathrm {a})·f^3(\mathrm {a})···f^{10}(\mathrm {a})=f^{1 +··· + 10}(\mathrm {a})=f^{55}(\mathrm {a})$?
∵?$f(\mathrm {a})=3$?
∴?$f^{55}(\mathrm {a})=3^{55}$?
