45. The domain of the function is $[0,13]$, because
$$\begin{aligned} y= & \sqrt{x+27}+\sqrt{13-x}+\sqrt{x}=\sqrt{x+27}+\sqrt{13+2 \sqrt{x(13-x)}} \geqslant \\ & \sqrt{27}+\sqrt{13}=3 \sqrt{3}+\sqrt{13} \end{aligned}$$

When $x=0$, the equality holds. Therefore, the minimum value of $y$ is $3 \sqrt{3}+\sqrt{13}$.
By the Cauchy-Schwarz inequality, we have
$$\begin{aligned} y^{2}= & (\sqrt{x+27}+\sqrt{13-x}+\sqrt{x})^{2} \leqslant \\ & \left(1+\frac{1}{3}+\frac{1}{2}\right)[(x+27)+3(13-x)+2 x]=121 \end{aligned}$$

Thus, $y \leqslant 11$. From the condition for equality in the Cauchy-Schwarz inequality, we get $4 x=9(13-x)=x+27$, solving which gives $x=9$.

Therefore, when $x=9$, the equality holds. Hence, the maximum value of $y$ is 11.