解：$(1) ②$已知$f(2)=5，$$f(2n)=25$
∵$f(2n)=f(2 + 2+···+2)(n$个$2$相加$)$
根据$f(\mathrm {m})·f(\mathrm {n})=f(m + n)，$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，$化简$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})$
∴$f(\mathrm {a})·f(2a)·f(3a)···f(10a)=f(\mathrm {a})·f^2(\mathrm {a})·f^3(\mathrm {a})···f^{10}(\mathrm {a})$
根据同底数幂相乘，底数不变，指数相加，$1 + 2+···+10=\frac {10×(1 + 10)}2=55$
则$f(\mathrm {a})·f^2(\mathrm {a})·f^3(\mathrm {a})···f^{10}(\mathrm {a})=f^{1 +··· + 10}(\mathrm {a})=f^{55}(\mathrm {a})$
∵$f(\mathrm {a})=3$
∴$f^{55}(\mathrm {a})=3^{55}$
