= Solution
{c}
Separate a long future count from the early <Poisson arrival time>. Let $\mathcal F_t=\sigma(N_s:0\leq s\leq t)$ be the <natural filtration>, and define
$$
A_n=\frac{N_{n^4}-N_{n^2}-(n^4-n^2)}{n^2},\qquad
B_n=\frac{\min(T_n,n^2)-n}{\sqrt n},\qquad
C_n=\frac{N_{n^2}-n^2}{n^2}.
$$
The truncated <Poisson arrival time> in $B_n$ is $\mathcal F_{n^2}$-measurable. The <independent increments> property makes $A_n$ independent of $\mathcal F_{n^2}$, and hence of $B_n$.
The numerator of $A_n$ is a centered <Poisson random variable> of parameter $n^4-n^2$. Apply the <central limit theorem> to the $n^4-n^2$ unit-time increments, then use $\sqrt{n^4-n^2}/n^2\to1$ and <Slutsky's theorem>. This proves $A_n\xrightarrow{\mathrm d}N(0,1)$. Also, <Markov's inequality> and $\mathbb ET_n=n$ give
$$
\mathbb P(T_n>n^2)\leq\frac1n\longrightarrow0.
$$
Thus $B_n$ equals $(T_n-n)/\sqrt n$ outside an event of vanishing <probability>, so part (i) and <Slutsky's theorem> give $B_n\xrightarrow{\mathrm d}N(0,1)$. Exact <independence of random variables> for each $n$ now gives the joint limit $(A_n,B_n)\xrightarrow{\mathrm d}(Z_1,Z_2)$ with independent <standard normal random variables>.
Finally, $\mathbb EC_n=0$ and $\operatorname{Var}(C_n)=n^{-2}$, so <Chebyshev's inequality> gives $C_n\to0$ in <probability>. The first coordinate in the question is $A_n+C_n$. The second differs from $B_n$ only on $\{T_n>n^2\}$, and so its difference tends to zero in <probability>. The vector version of <Slutsky's theorem> proves the required joint <convergence in distribution>. The boundary of $(-\infty,x]\times[y,\infty)$ has zero limiting <probability>, since both <marginal distributions> have <continuous> <cumulative distribution functions>. The <Portmanteau theorem> therefore yields
$$
\boxed{\mathbb P\!\left(\frac{N_{n^4}-n^4}{n^2}\leq x,\frac{T_n-n}{\sqrt n}\geq y\right)\longrightarrow\Phi(x)\bigl(1-\Phi(y)\bigr).}
$$
This is $\mathbb P(Z_1\leq x,Z_2\geq y)$ by <independence of random variables>.
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