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Answer 1

$\frac{a^2-25}{a+3} \cdot\frac{1}{a^2+5a} -\frac{a^2+10a+25}{a^3-6a^2+9a} :\frac{a+5}{a-3}+(\frac{4}{a-3})^{2} = \frac{(a-5)(a+5)}{a+3} \cdot\frac{1}{a(a+5)} -\frac{(a+5)^2}{a(a^2-6a+9)} :\frac{a+5}{a-3}+(\frac{4}{a-3})^{2}=\frac{a-5}{a(a+3)} -\frac{(a+5)^2}{a(a-3)^2} \cdot\frac{a-3}{a+5}+(\frac{4}{a-3})^{2}=\frac{a-5}{a(a+3)} -\frac{a+5}{a(a-3)} +\frac{16}{(a-3)^2}=\frac{(a-5)(a-3)-(a+5)(a+3)}{a(a+3)(a-3)} +\frac{16}{(a-3)^2}=\frac{a^2-8a+15-a^2-8a-15}{a(a+3)(a-3)} +\frac{16}{(a-3)^2}$$=\frac{-16a(a-3)}{a(a+3)(a-3)} +\frac{16a(a+3)}{(a-3)^2}=\frac{-16a^2+48a+16a^2+48a}{a(a+3)(a-3)^2} =\frac{96a}{a(a+3)(a-3)^2} =\frac{96}{(a+3)(a-3)^2}.$