If $I$ is not compact, the claim is false:
Let $I=(0,1)$ and $F_n(x)=x^n$. Then $F_n$ is a strictly increasing function. and $(F_n)$ converges pointwise to the continuous and bounded zero function. However, the convergence is not uniform as $\sup |F_n(x)-F(x)|=1$ for all $n$.
Now assume that $I=[a,b]$ is compact. (Thus the condition that $F$ be bounded is superfluous: It is a consequence of its continuity).
Assume $\epsilon>0$ is given.
For $x\in I$, the set $U_x:=F^{-1}\left((F(x)-\frac\epsilon9,F(x)+\frac\epsilon9)\right)$ is a relative open subset of $I$.
Hence we can find $r_x>0$ such that the relative open set $V_x:=B(x,r_x)\cap I$ is $\subseteq U_x$.
We have $I=\bigcup_{x\in I} V_x$.
By compactness this there exists a finite subcover, i.e. we find $x_0<x_1<\cdots <x_m$ such that $I=\bigcup_{x\in I} V_x$.
Wlog. $x_0=a$, $x_m=b$.
Among all such sequences $(x_k)$ we select one with minimal $m$.
Assume there is a $k$ such that $V_{x_k}\cap V_{x_{k+1}}=\emptyset$.
Then there is a point $x$ between $V_{x_k}$ and $V_{x_{k+1}}$ (we may take for example $x=x_k+r_k$) that is covered by a $V_{x_i}$ with either $i<k$ or $i>k+1$. In the first case, we see that $V_{x_k}\subset V_{x_i}$ and hence $x_k$ can be dropped; in the second case, we can similarly drop $x_{k+1}$. In both cases we obtain a shorter sequence, contrary to the assumption that $m$ is minimal (note that we do not drop $x_0$ or $x_m$).
From $V_{x_k}\cap V_{x_{k+1}}\ne\emptyset$ we conclude that $|F(x_k)-F(x_{k+1})|<\frac{2\epsilon}9$ (because $|F(x_k)-F(x)|<\frac\epsilon9$ and $|F(x_{k+1})-F(x)|<\frac\epsilon9$ for some $x\in V_{x_k}\cap V_{x_{k+1}}$).
In fact, $x_k\le x\le x_{k+1}$ implies that $|F(x)-F(x_k)|<\frac\epsilon9$ or $|F(x)-F(x_{k+1})|<\frac\epsilon9$ (in other words: $|F(x)-F(x_{r})|<\frac\epsilon9$ for $r=k$ or $r=k+1$).
Per pointwise convergence at $x_0,\ldots,x_m$ there exists $N\in\mathbb N$ such that $|F_n(x_k)-F(x_k)|<\frac{2\epsilon}9$ for all $n>N$ and $0\le k\le m$.
Then for $n>N$ and $x\in I$ we find $k$ with $x_k\le x\le x_{k+1}$.
Then
$$|F_n(x_{k+1})-F_n(x_k)|\\\le |F_n(x_{k+1})-F(x_{k+1})|+|F(x_{k+1})-F(x_k)| +|F(x_{k})-F_n(x_k)|\\
<\frac{2\epsilon}9+\frac{2\epsilon}9+\frac{2\epsilon}9=\frac{2\epsilon}3.$$
By the above remarks we have $|F(x)-F(x_r)|<\frac\epsilon9$ for $r=k$ or $r=k+1$.
and therefore (using either $F_n(x_k)=F_n(x_r)\le F_n(x)\le F_n(x_{k+1})$ or $F_n(x_k)\le F_n(x)\le F_n(x_r)=F_n(x_{k+1})$)
$$|F_n(x)-F(x)|\le |F_n(x)-F_n(x_r)|+|F_n(x_r)-F(x_r)|+|F(x_r)-F(x)|\\
<|F_n(x_{k+1})-F_n(x_k)|+\frac{2\epsilon}9+\frac\epsilon9\\
<\frac{2\epsilon}3+\frac{2\epsilon}9+\frac\epsilon9=\epsilon. $$
Therefore $\sup |F_n(x)-F(x)|<\epsilon$ for all $n>N$, i.e. the convergence is uniform.
Hints:
(a) A function is Riemann integrable if
$\quad(i)$ it is bounded, and
$\quad(ii)$ has a countable set of discontinuities.
You can see also Lebesgue criteria for Riemann integrability.
(c) you need the fact that if $g$ is a continuous, then
$$ x_n \to x \implies g(x_n)\to g(x)\quad \mathbb{as}\quad n\to \infty, $$
and notice that
$$ |f_n(x_n)-f(x)|=|(f_n(x_n)-f_n(x))+(f_n(x)-f(x))|$$
$$\leq |f_n(x_n)-f_n(x)|+|f_n(x)-f(x)| < \dots.$$
Best Answer
(a) is true: $\sup_{x\in[0,1]}|f_n(x)-f(x)|=\sup_{x\in[0,1]}|f_n(x)|=f_n(1)\to0$
(b) is false: Consider for $n\in\mathbb N,~f_n:[0,1]\to\mathbb R:x\mapsto$$ \begin{cases} 0, & \text{if}~0\leq x\leq1-\frac{1}{n} \\ \frac{1}{n}, & \text{if}~1-\frac{1}{n}<x\leq 1 \\ \end{cases}$