Answer
(a) $F=\frac{\mu_s mg}{cos\theta-\mu_s sin\theta}$
(b) Please see the work below.
Work Step by Step
(a) We can find the required force as follows:
We know that $\Sigma F_y=0$
$\implies -Fsin\theta-mg+N=0$
$\implies N=Fsin\theta+mg$......eq(1)
Now $f_s=\mu_sN$
$\implies f_s=\mu_s(Fsin\theta+mg)$....eq(2)
Similarly $\Sigma F_x=0$
$Fcos\theta-f_s=0$
$\implies Fcos\theta=f_s$......eq(3)
Comparing eq(2) and (3), we obtain:
$Fcos\theta=\mu_s(Fsin\theta+mg)$
$\implies F(cos\theta-\mu_s sin\theta)=\mu_s mg$
This simplifies to:
$F=\frac{\mu_s mg}{cos\theta-\mu_s sin\theta}$
(b) For $\mu_s=\frac{1}{tan\theta}$
$\implies \mu_s=\frac{cos\theta}{sin\theta}$
$F=\frac{(\frac{cos\theta}{sin\theta})mg}{cos\theta-(\frac{cos\theta}{sin\theta})sin\theta}$
$\implies F=\frac{(cos\theta) mg}{sin\theta(cos\theta-cos\theta)}$
$\implies F=\frac{(cos\theta) mg}{0}$
Thus, the applied force becomes infinite and therefore it is impossible to move the crate. The same result is obtained for $\mu_s\gt \frac{1}{tan\theta}$.
