My attempt:
i) $$\lim_{t\to\ 0} \int_t^1 \ x^{-0.5}\ $$ = $$\lim_{t\to\ 0} 2(1^{0.5}) - 2(t^{0.5}) = 2 - 0 = 2 $$ (since $lim_{t\to\ 0^+} t^{0.5} = lim_{t\to\ 0^-} t^{0.5} = 0 $)
ii) $$\lim_{t\to\ ∞} \int_1^∞ \ x^{-0.5}\ $$ = $$\lim_{t\to\ ∞} 2\ ∞^{0.5}\ -\ 1^{0.5}\ = ∞ $$
I know that this answer is incorrect (http://www.wolframalpha.com/input/?i=integral+from+1+to+infinity+%28x%5E-0.5%29), so hopefully somebody call tell me where I went wrong.
iii) f is not improperly Riemann Integrable over [0,∞) because $\frac{1}{\sqrt{0}}$ involves division by zero and so does not exist.
iv) f is not improperly integrable over {-1,1} because f is not continuous at x = 0, and therefore not continuous on the interval {-1,1}
Let's talk about the first one. This function has a singularity at zero. That means that for it to be improperly Riemann integrable over $[0,1]$ you need
$$\lim_{a \to 0^+} \int_a^1 \frac{1}{\sqrt{x}} dx$$
to exist as a finite number. Using the power rule you find that this integral is $2-2\sqrt{a}$, which actually does converge as $a \to 0^+$. So that covers the first part.
That's the situation with a bounded interval. With an unbounded interval you also have to deal with taking the unbounded limit to infinity. That is, $\int_1^\infty \frac{1}{\sqrt{x}} dx = \lim_{b \to \infty} \int_1^\infty \frac{1}{\sqrt{x}} dx$. Try to compute this the same way as you did before and see if you get a finite number. That covers the second part.
If you can do the first and second parts then you can do the third and fourth parts. (In the fourth part you will need to split the interval in two pieces.)