jmc

Since $p(2)=p(-1)=0,$ $p(x)$ is of the form $$p(x)=(ax+b)(x-2)(x+1)$$for some constants $a$ and $b.$ Setting $x=4$ and $x=5,$ we get $$\begin{array}{rl}(4a+b)(2)(5)& =p(4)=6,\\ (5a+b)(3)(6)& =p(5)=8,\end{array}$$so $$\begin{array}{rl}4a+b& =\frac{3}{5},\\ 5a+b& =\frac{4}{9}.\end{array}$$Solving, we find $a=-\frac{7}{45}$ and $b=\frac{11}{9}.$ Hence, $$p(x)=\left(-\frac{7}{45}x+\frac{11}{9}\right)(x-2)(x+1)=-\frac{(7x-55)(x-2)(x+1)}{45}.$$Therefore, $$p(7)=-\frac{(49-55)(5)(8)}{45}=\boxed{\frac{16}{3}}.$$