1.x = log3(36)= 2 log3(6)= 2lg 6/(lg3)；

y = log4(36)= 2 log4(6)= log2(6)= lg6/lg2 .

therefore

(2/x)+( 1/y)

= monotonically increasing,

The minimum value is h (a) = g (1/3) = (28/9)-(2a/3);

When 1/3

h(a)=g(a)=3-a^2；

When a & gt=3, g(x) decreases monotonically in this interval,

The minimum value is h(a)=g(3)= 12-6a.

(2) By (1)

h(a)=(28/9)-(2a/3)(a & lt； = 1/3)

h(a)=3-a^2( 1/3 & lt； a & lt3)

h(a)= 12-6a(a & gt； =3)

Let such m and n exist and satisfy m >；; n & gt3。

When a∈[n, m],

The function h(a)= 12-6a is a monotonically decreasing linear function,

The maximum value is h (n) =12-6n;

The minimum value is h(m)= 12-6m.

therefore

12-6n=m^2

12-6m=n^2

Solve m and n at the same time, and then make a choice according to the value range (3, positive infinity).

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