= Solution
Write $S_N=\sum_{i=1}^NT_i$. The answer is
$$
\boxed{\text{speed }N,\qquad I(x)=
\begin{cases}\lambda x,&x\geq0,\\+\infty,&x<0.\end{cases}}
$$
The <sum of exponential variables with linearly increasing rates> has the distribution of the maximum of $N$ <independent> rate-$\lambda$ <exponential random variables>. To prove this identity, start $N$ <independent> rate-$\lambda$ exponential clocks. The first rings after a rate-$N\lambda$ exponential time; the <memoryless property> leaves $N-1$ fresh <independent> clocks. Repeating this argument shows that the successive <order statistic> gaps are <independent>, with rates $N\lambda,(N-1)\lambda,\ldots,\lambda$. Their sum has the distribution of $S_N$. Therefore
$$
\mathbb P(S_N\leq t)=(1-e^{-\lambda t})^N\quad(t\geq0).
$$
For $x>0$, $N e^{-\lambda Nx}\to0$, and hence
$$
\mathbb P(S_N>Nx)=1-(1-e^{-\lambda Nx})^N
\sim N e^{-\lambda Nx},\qquad
\lim_NN^{-1}\log\mathbb P(S_N>Nx)=-\lambda x.
$$
If $0<u<v$, subtracting the two tails gives the interval exponent $-\lambda u$, since the tail at $Nv$ is exponentially smaller. Also $\mathbb E S_N=\lambda^{-1}\sum_{i=1}^N i^{-1}=O(\log N)$, so the <Markov inequality> gives $S_N/N\to0$ in <probability>. Local lower bounds at positive points and at zero follow; the tail estimate gives the closed-set upper bound exactly as in part (a). This proves the full <large deviation principle>.
There is also a deduction from part (b). For fixed $k$ and $i>k$, a rate-$i\lambda$ <exponential random variable> is stochastically smaller than a rate-$k\lambda$ one. Coupling by the same uniform variables therefore gives $S_N\leq S_k+Z_N$ with <independent> additional copies. For fixed $x>0$ and sufficiently large $k$, the upper-tail exponent from part (b) is $-\lambda x-\log((k-1)/k)$. Let $k\to\infty$ to obtain $-\lambda x$; the lower tail estimate $\mathbb P(S_N>Nx)\geq\mathbb P(T_1>Nx)=e^{-\lambda Nx}$ matches it. The exact maximum identity additionally supplies all local lower bounds.
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