(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})$
$Given V= \large\frac{3}{4}v_e=\large\frac{3}{4}\sqrt{\large\frac{2GM}{R}}$
$=\large\frac{1}{2}m \times \large\frac{9}{16} \times (\large\frac{2GM}{R})+(\large\frac{-GMm}{R})$
$=(\large\frac{-GMm}{R+h})$
$\large\frac{9GMm}{16R}-\large\frac{GMm}{R}=-\large\frac{GMm}{R+h}$
$\large\frac{-7GMm}{16R}=-\large\frac{GMm}{R+h}$
$\large\frac{7}{16R}=\large\frac{1}{R+h}$
$\Rightarrow 7R+7h=16R$
$h=\large\frac{9}{7}R$