\[
\mathbf{F}(x, y, z)=\frac{\sqrt{x}}{6+z} \mathbf{i}+\frac{\sqrt{y}}{6+x} \mathbf{j}+\frac{\sqrt{z}}{6+y} \mathbf{k}
\]
(a) Find the curl of the vector field.
\[
\operatorname{curl}(F)=
\]
(b) Find the divergence of the vector field.
\[
\operatorname{div}(F)=
\]

Step 1 ：Given the vector field \(\mathbf{F}(x, y, z)=\frac{\sqrt{x}}{6+z} \mathbf{i}+\frac{\sqrt{y}}{6+x} \mathbf{j}+\frac{\sqrt{z}}{6+y} \mathbf{k}\)

Step 2 ：We can find the curl of the vector field by calculating the determinant of a special matrix that includes the unit vectors i, j, k in the first row, the partial derivatives with respect to x, y, and z in the second row, and the vector field components in the third row.

Step 3 ：\(F_i = \frac{\sqrt{x}}{z + 6}\), \(F_j = \frac{\sqrt{y}}{x + 6}\), \(F_k = \frac{\sqrt{z}}{y + 6}\)

Step 4 ：The curl of the vector field F is given by the matrix \(\begin{bmatrix} 0 \\ \frac{\sqrt{x}}{(z + 6)^2} \\ -\frac{\sqrt{y}}{(x + 6)^2} + \frac{\sqrt{z}}{(y + 6)^2} \end{bmatrix}\)

Step 5 ：We can find the divergence of the vector field by summing the partial derivatives of each component of the field with respect to its corresponding variable.

Step 6 ：\(F_i = \frac{\sqrt{x}}{z + 6}\), \(F_j = \frac{\sqrt{y}}{x + 6}\), \(F_k = \frac{\sqrt{z}}{y + 6}\)

Step 7 ：The divergence of the vector field F is \(\frac{1}{2\sqrt{z}(y + 6)} + \frac{1}{2\sqrt{y}(x + 6)} + \frac{1}{2\sqrt{x}(z + 6)}\)

Step 8 ：Final Answer: The curl of the vector field F is \(\boxed{\begin{bmatrix} 0 \\ \frac{\sqrt{x}}{(z + 6)^2} \\ -\frac{\sqrt{y}}{(x + 6)^2} + \frac{\sqrt{z}}{(y + 6)^2} \end{bmatrix}}\) and the divergence of the vector field F is \(\boxed{\frac{1}{2\sqrt{z}(y + 6)} + \frac{1}{2\sqrt{y}(x + 6)} + \frac{1}{2\sqrt{x}(z + 6)}}\)