Answer: \(a=2\), \(b=-1\).

Since \(x+2\) is a factor of \(\mathrm p(x)\), the factor theorem gives

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

Substitute \(x=-2\) into \(\mathrm p(x)=x^3+ax^2+bx-2\):

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

So

\(-8+4a-2b-2=0,\)

which simplifies to

\(4a-2b=10.\)

Divide by \(2\):

\(2a-b=5. \tag{1}\)

By the remainder theorem, the remainder when \(\mathrm p(x)\) is divided by \(x-3\) is \(\mathrm p(3)\). This remainder is \(40\), so

\(3^3+a(3)^2+b(3)-2=40.\)

Therefore

\(27+9a+3b-2=40,\)

so

\(9a+3b=15.\)

Divide by \(3\):

\(3a+b=5. \tag{2}\)

From (1),

\(b=2a-5.\)

Substitute this into (2):

\(3a+(2a-5)=5.\)

So

\(5a=10,\)

giving

\(a=2.\)

Then

\(b=2(2)-5=-1.\)

Therefore

\(a=2,\qquad b=-1.\)