Step-by-step Solution:

Correct Answer: B $\left(\sqrt{\frac{7}{3}},\,2\right)$ This problem requires finding the range of eccentricity by using the geometric properties of a hyperbola. Step 1: Determine the Coordinates of Key Points Given the standard hyperbola \[ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1, \] the foci are \(F_1=(-ae,0)\) and \(F_2=(ae,0)\), where \(e\) is the eccentricity. Point \(M\): The condition \(MF_2 \perp F_1F_2\) means that the segment \(MF_2\) is perpendicular to the \(x\)-axis. Hence \(M\) lies on the latus rectum through \(F_2\), so the \(x\)-coordinate of \(M\) is \(ae\). Substitute \(x=ae\) in the hyperbola equation: \[ \frac{(ae)^2}{a^2}-\frac{y^2}{b^2}=1 \;\Rightarrow\; e^2-\frac{y^2}{b^2}=1 \;\Rightarrow\; y^2=b^2(e^2-1). \] Using \(b^2=a^2(e^2-1)\), we get \[ y^2=a^2(e^2-1)^2. \] Since \(M\) is above the \(x\)-axis, \[ y=a(e^2-1). \] So, \[ M\bigl(ae,\;a(e^2-1)\bigr). \] Step 2: Use the Perpendicularity Condition \(MF_1 \perp OH\) The condition \(MF_1 \perp OH\) means that \(H\) is the foot of the perpendicular from the origin \(O\) to the line \(MF_1\). Thus \(|OH|\) is the perpendicular distance from \((0,0)\) to the line \(MF_1\). Equation of the line \(MF_1\): Points are \(M(ae,a(e^2-1))\) and \(F_1(-ae,0)\). Slope: \[ m=\frac{a(e^2-1)-0}{ae-(-ae)}=\frac{a(e^2-1)}{2ae}=\frac{e^2-1}{2e}. \] Line through \(F_1(-ae,0)\): \[ y=\frac{e^2-1}{2e}(x+ae). \] Rearranging: \[ (e^2-1)x-2ey+ae(e^2-1)=0. \] Distance \(|OH|\) from origin to \(Ax+By+C=0\) is \(\dfrac{|C|}{\sqrt{A^2+B^2}}\). Hence \[ |OH|=\frac{|ae(e^2-1)|}{\sqrt{(e^2-1)^2+(-2e)^2}} =\frac{ae(e^2-1)}{\sqrt{e^4-2e^2+1+4e^2}} =\frac{ae(e^2-1)}{\sqrt{e^4+2e^2+1}} =\frac{ae(e^2-1)}{\sqrt{(e^2+1)^2}} =\frac{ae(e^2-1)}{e^2+1}. \] Step 3: Relate Eccentricity \(e\) to \(\lambda\) Given \(|OH|=\lambda|OF_2|\). Since \(F_2(ae,0)\), \[ |OF_2|=ae. \] Substitute: \[ \frac{ae(e^2-1)}{e^2+1}=\lambda(ae). \] Cancel \(ae\) (as \(a>0\) and \(e>1\)): \[ \lambda=\frac{e^2-1}{e^2+1}. \] Step 4: Find the Range of Eccentricity \(e\) Given \(\lambda\in\left(\frac{2}{5},\frac{3}{5}\right)\), so \[ \frac{2}{5}<\frac{e^2-1}{e^2+1}<\frac{3}{5}. \] Left inequality: \[ \frac{2}{5}<\frac{e^2-1}{e^2+1} \Rightarrow 2(e^2+1)<5(e^2-1) \Rightarrow 2e^2+2<5e^2-5 \Rightarrow 7<3e^2 \Rightarrow e^2>\frac{7}{3} \Rightarrow e>\sqrt{\frac{7}{3}}. \] Right inequality: \[ \frac{e^2-1}{e^2+1}<\frac{3}{5} \Rightarrow 5(e^2-1)<3(e^2+1) \Rightarrow 5e^2-5<3e^2+3 \Rightarrow 2e^2<8 \Rightarrow e^2<4 \Rightarrow e<2. \] Combining: \[ e\in\left(\sqrt{\frac{7}{3}},\,2\right). \]