$\begin {array} {1 1} (1)\;\large\frac{9}{7}R & \quad (2)\;\large\frac{9}{8}R \\ (3)\;\large\frac{10}{3}R & \quad (4)\;\large\frac{10}{9}R \end {array}$

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(1) $\large\frac{9}{7}R$

Explanation:

According to the law of conservation of energy,

$(T.E)_{at\;surface} = (T.E)_{at\;height}$

$(K.E+P.E)_{surface} =(K.E+P.E)_{max \;height}$

$\large\frac{1}{2}mv^2 +(\large\frac{-GMm}{R})$ =$ 0+(\large\frac{-GMm}{R+h})$

$Given V= \large\frac{3}{4}v_e=\large\frac{3}{4}\sqrt{\large\frac{2GM}{R}}$

$=\large\frac{1}{2}m \times \large\frac{9}{16} \times (\large\frac{2GM}{R})+(\large\frac{-GMm}{R})$

$=(\large\frac{-GMm}{R+h})$

$\large\frac{9GMm}{16R}-\large\frac{GMm}{R}=-\large\frac{GMm}{R+h}$

$\large\frac{-7GMm}{16R}=-\large\frac{GMm}{R+h}$

$\large\frac{7}{16R}=\large\frac{1}{R+h}$

$\Rightarrow 7R+7h=16R$

$h=\large\frac{9}{7}R$

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