Answer: \(n=6\); \(a=\frac1{16}\), \(c=\frac{15}{16}\).

The constant term in \((2+ax)^n\) is \(2^n\). Since this is \(64\),

\(2^n=64,\)

so \(n=6\).

Now expand the first three terms of \((2+ax)^6\):

\((2+ax)^6=2^6+{6\choose1}2^5(ax)+{6\choose2}2^4(ax)^2+\cdots.\)

Thus

\(b=6\cdot 2^5a=192a\)

and

\(c=15\cdot 2^4a^2=240a^2.\)

Therefore

\(5b^2=5(192a)^2=184320a^2\)

and

\(768c=768(240a^2)=184320a^2.\)

So

\(5b^2=768c.\)

If \(b=12\), then

\(192a=12,\)

so

\(a=\frac1{16}.\)

Then