Answer: \(a=2\), \(b=0\).

Since \(x-3\) is a factor of \(\mathrm{p}(x)\), the factor theorem gives

\(\mathrm{p}(3)=0\).

Substitute \(x=3\) into

\(\mathrm{p}(x)=ax^3-7x^2-bx+9\):

\(27a-7(9)-3b+9=0\).

This simplifies to

\(27a-3b-54=0\).

Dividing by \(3\),

\(9a-b=18\). \(\quad\text{(1)}\)

When \(\mathrm{p}(x)\) is divided by \(x+2\), the remainder is \(\mathrm{p}(-2)\). The remainder is given as \(-35\), so

\(\mathrm{p}(-2)=-35\).

Substituting \(x=-2\),

\(-8a-7(4)+2b+9=-35\).

So

\(-8a+2b-19=-35\),

which gives

\(-8a+2b=-16\).

Dividing by \(2\),

\(-4a+b=-8\). \(\quad\text{(2)}\)

Add equations (1) and (2):

\((9a-b)+(-4a+b)=18-8\).

Thus

\(5a=10\),

so

\(a=2\).

Substitute into \(-4a+b=-8\):

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

so

\(b=0\).