The given joint distribution is: $$ f_{X, Y}(x, y)=\frac{n 2 a^2 x^{n-1}}{y^{n+3}} I_{[0, y]}(x) I_{(a, \infty)}(y), n \in \mathbb{N} $$ The exercise asks us to find the cumulative distribution of the conditional $X \mid Y=y, y>a$
$$ f_{X \mid Y:}(x \mid y)=\frac{f_{X, Y}(x, y)}{f_Y(y)} $$
So we need to calculate the marginal distribution of $Y$:
$$ \begin{aligned} f_Y(y) & =\int_{-\infty}^{\infty} f_{X, Y}(x, y) d x \\ & =\int_0^y y^{-(n+3)} \cdot n 2 a^2 x^{n-1} I_{(a, \infty)}(y) d x \\ & =y^{-(n+3)} n 2 a^2 I_{(a, \infty)}(y) \int_0^y x^{n-1} d x \\ & =\left.y^{-(n+3)} n 2 a^2 I_{(u, \infty)}(y) \frac{x^n}{n}\right|_0 ^y \\ & =y^{-n+3} \cdot y^n 2 a^2 I_{(a, \infty)}(y) \\ & =\frac{2 a^2}{y^3} I_{(a, \infty)}(y) \end{aligned} $$ And we can finally $$ \begin{aligned} & f_{X \mid Y}(x \mid y)=\frac{n 2 a^2 x^{n-1}}{y^{n+3}} I_{(0, y]}(x) I_{(a, \infty )}(y) \cdot \frac{y^3}{2 a^2 I_{(a, \infty)}(y)} \\ & =\frac{n x^{n-1}}{y^n} I_{[0, y]}(x), y>a \\ & \end{aligned} $$ We are allowed to divide by the indicator function because we are in a region where it is equal to 1 only. Now that we have the distribution function we just need to calculate the cumulative $$ \begin{aligned} F_{X \mid Y}(t \mid y) & =\int_0^t \frac{n x^{n-1}}{y^n} d x= \\ & =\frac{n}{y^n} \int_0^t \frac{x^{n-1}}{} d x \text {, se } 0 \leqslant t<y \\ & =\left.\frac{n}{y^n} \frac{1}{n} x^n\right|_0 ^t=\frac{t^n}{y^n}, 0 \leqslant t<y \end{aligned} $$ Precisely, with the segments of the real line properly defined, we can build the full cumulative distribution function as:
$$ F_{X \mid Y}(t \mid y)=\left\{\begin{array}{lll} 0 & \text { se } & t<0 \\ \frac{t^n}{y^n} & \text { se } & 0 \leqslant t \leqslant y \\ 1 & \text { se } & t \geqslant y \end{array}\right. $$
For $y>a$.
Finally, the exercise asks us to take the limit of $F_{X \mid Y}$ as $n$ goes to infinity and identify what is the resulting distribution. But I can't identify any, leading me to believe I made a mistake somewhere (or the question is ill-posed, which I doubt).