Answer
$a)$ There is no indeterminate form obtained from direct substitution.
$b)$ $0$
$c)$ see picture
Work Step by Step
$a)$ $\lim\limits_{x \to 0^{+}}x^{\frac{1}{x}}=(0^{+})^{\frac{1}{0^{+}}}=(0^{+})^{\infty}$
$0^{\infty}$ is not an indeterminate form.
$b)$ $\lim\limits_{x \to 0^{+}}x^{\frac{1}{x}}=(0^{+})^{\frac{1}{0^{+}}}=(0^{+})^{\infty}=0$
$c)$ see picture