Answer: \(1537024\).

First expand the second factor:

\((1-4x^2)^2=1-8x^2+16x^4\).

Now consider

\(\left(2+\frac{3}{x^2}\right)^{10}\).

The general term in this expansion is

\(\binom{10}{r}2^{10-r}\left(\frac{3}{x^2}\right)^r=\binom{10}{r}2^{10-r}3^r x^{-2r}\).

To get a constant term after multiplying by \(1-8x^2+16x^4\), we only need the terms in \(x^0\), \(x^{-2}\) and \(x^{-4}\).

These correspond to \(r=0,1,2\):

\(r=0:\quad 2^{10}=1024\),

\(r=1:\quad \binom{10}{1}2^9\cdot3x^{-2}=15360x^{-2}\),

\(r=2:\quad \binom{10}{2}2^8\cdot3^2x^{-4}=103680x^{-4}\).

The constant term is obtained from

\(1024(1)+(15360x^{-2})(-8x^2)+(103680x^{-4})(16x^4)\).

Therefore the required term is

\(1024-122880+1658880=1537024\).

Answer: \(a=3\), \(b=\pm\sqrt{6720}=\pm8\sqrt{105}\).

The expansion is

\(5(2+ax)^n.\)

The constant term is

\(5\cdot2^n.\)

This is given as \(640\), so

\(5\cdot2^n=640.\)

Thus

\(2^n=128=2^7,\)

so

\(n=7.\)

Now consider the coefficient of \(x^2\). In \((2+ax)^7\), the \(x^2\) term is

\(\binom72 2^5(ax)^2.\)

After multiplying by \(5\), the coefficient of \(x^2\) is

\(5\binom72 2^5a^2.\)

Since this is \(30240\),

\(5\cdot21\cdot32\cdot a^2=30240.\)

So

\(3360a^2=30240,\)

and hence

\(a^2=9.\)

Therefore \(a=3\) or \(a=-3\).

The coefficient of \(x\) is

\(5\binom71 2^6a=5\cdot7\cdot64a=2240a.\)

This coefficient is given as \(b^2\), which must be non-negative. If \(a=-3\), the coefficient \(2240a\) would be negative, impossible for \(b^2\). Therefore

\(a=3.\)

Then

\(b^2=2240(3)=6720.\)

Hence

\(b=\pm\sqrt{6720}=\pm8\sqrt{105}.\)

Answer: (a) \(\binom{59}{r}2^{59-r}\). (b) \(r=19\).

(a) In the expansion of \((2+x)^{59}\), the general term containing \(x^r\) is

\(\binom{59}{r}2^{59-r}x^r.\)

Therefore the coefficient of \(x^r\) is

\(\binom{59}{r}2^{59-r}.\)

(b) The coefficient of \(x^{r+1}\) is

\(\binom{59}{r+1}2^{59-(r+1)}=\binom{59}{r+1}2^{58-r}.\)

Set the two coefficients equal:

\(\binom{59}{r}2^{59-r}=\binom{59}{r+1}2^{58-r}.\)

Divide both sides by \(\binom{59}{r}2^{58-r}\):

\(2=\dfrac{\binom{59}{r+1}}{\binom{59}{r}}.\)

Now

\(\dfrac{\binom{59}{r+1}}{\binom{59}{r}}=\dfrac{59-r}{r+1}.\)

So

\(2=\dfrac{59-r}{r+1}.\)

Multiplying by \(r+1\),

\(2r+2=59-r.\)

Thus

\(3r=57,\)

and hence

\(r=19.\)