Didn't include my drafting but not sure if my answer is right.
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1Just a heads up: this question might not be received very well by the community since you haven't typed your question using MathJax and you haven't shown any of your work. The community at large generally responds positively to questions that also include explanations of how you arrived at your answer(s). – Clayton May 18 '19 at 14:26
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In e) I think it should be $a>-2$. Otherwise I don't see problems. – Jakobian May 18 '19 at 14:36
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@Jakobian Riemann integrable implies bounded – user10354138 May 18 '19 at 14:38
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Agreed. When we talk about the usual definition. I was thinking more in the terms of improper integrals. – Jakobian May 18 '19 at 14:40
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@Jakobian In e), could you please explain why you reached a>-2. Thanks:) – Jimmy May 19 '19 at 18:22
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@Jimmy Like user pointed out, I think you need to use the standard definition of a Riemann integral, not improper integrals. But I might explain how I did it. $\int_0^1 x^a\cos(1/x)\ dx = \int_1^\infty \frac{1}{u^{a-2}}\cos(u)\ du$. From Dirichlet test, the integral is convergent when $a>2$. Otherwise it's obviously divergent. – Jakobian May 19 '19 at 21:09