Answer: \(a=3\), \(b=-4\).

We start with the main method. Use the binomial theorem to identify only the terms needed, rather than expanding the whole expression.

First expand the terms needed from \((2+ax)^5\):

\((2+ax)^5=32+80ax+80a^2x^2+\cdots.\)

Now multiply by \((1+bx)\):

\((2+ax)^5(1+bx)=(32+80ax+80a^2x^2+\cdots)(1+bx).\)

The coefficient of \(x\) is

\(80a+32b.\)

Since the given coefficient of \(x\) is \(112\),

\(80a+32b=112.\)

Divide by \(16\):

\(5a+2b=7.\)

The coefficient of \(x^2\) is

\(80a^2+80ab.\)

Since the given coefficient of \(x^2\) is \(-240\),

\(80a^2+80ab=-240.\)

Divide by \(80\):

\(a^2+ab=-3.\)

From \(5a+2b=7\),

\(b=\frac{7-5a}{2}.\)

Substitute this into \(a^2+ab=-3\):

\(a^2+a\left(\frac{7-5a}{2}\right)=-3.\)

Multiply by \(2\):

\(2a^2+7a-5a^2=-6.\)

So

\(3a^2-7a-6=0.\)

Factorising,

\((3a+2)(a-3)=0.\)

Since \(a\) is an integer,

\(a=3.\)

Then

\(5(3)+2b=7,\)

so

\(b=-4.\)

This completes the solution and gives the required result.