(a) \[p(x)=(3x+2)(3x^2+4)+(k-8).\] Multiplying the quotient \(3x^2+4\) gives \(9x^3+6x^2+12x+8\), leaving remainder \(k-8\).

(b) \[\int_1^6\frac{p(x)}{3x+2}\,dx=\left[x^3+4x+\frac{k-8}{3}\ln(3x+2)\right]_1^6=235+\frac{k-8}{3}\ln4.\] Comparing with \(a+\ln64= a+3\ln4\), gives \(a=235\) and \((k-8)/3=3\), so \(k=17\).