Answer: \(a=-3\), \(b=2\); remainder \(4\).

Since \(\mathrm p(x)\) is divisible by \(x-2\),

\(\mathrm p(2)=0.\)

So

\(8a+4b+12+4=0,\)

which simplifies to

\(2a+b=-4.\)

Now

\(\mathrm p'(x)=3ax^2+2bx+6.\)

When \(\mathrm p'(x)\) is divided by \(x+1\), the remainder is \(\mathrm p'(-1)\). Hence

\(3a-2b+6=-7,\)

so

\(3a-2b=-13.\)

Using \(b=-4-2a\) in \(3a-2b=-13\),

\(3a-2(-4-2a)=-13.\)

Therefore

\(7a=-21,\)

so \(a=-3\). Then

\(b=-4-2(-3)=2.\)

Using these values,

\(\mathrm p''(x)=6ax+2b=-18x+4.\)

The remainder when \(\mathrm p''(x)\) is divided by \(x\) is \(\mathrm p''(0)\), which is

\(4.\)