(a) The differential equation is \(\frac{dy}{dx} = k \frac{y}{x\sqrt{x}}\). Separate variables: \(\frac{dy}{y} = k \frac{dx}{x\sqrt{x}}\). Integrate both sides: \(\ln y = -2k \frac{1}{\sqrt{x}} + C\). Use the point \((1, 1)\) to find \(C\): \(\ln 1 = -2k \cdot 1 + C\), so \(C = 2k\). Use the point \((4, e)\) to find \(k\): \(\ln e = -2k \cdot \frac{1}{2} + 2k\), solving gives \(k = 1\). Thus, \(y = \exp\left(-\frac{2}{\sqrt{x}} + 2\right)\).

(b) As \(x\) tends to infinity, \(-\frac{2}{\sqrt{x}}\) tends to 0, so \(y\) approaches \(e^2\).