To deal with first point, it's nesessary to know what definition of independence of two random values you can use.
Say, for the following definition:
$X$ is idependent on $X$, iff for any $x,y\in\mathbb R$
$$\mathbb P(X\leq x, X\leq y)=\mathbb P(X\leq x)\mathbb P(X\leq y).$$
Таke $x=y$ be any real number. Then
$$\mathbb P(X\leq x)=\left(\mathbb P(X\leq x)\right)^2,$$
that is CDF $F_X(x)=\mathbb P(X\leq x)$ is either $0$ or $1$. Since it is non-decreasing and have zero limit at $-\infty$, unit limit at $+\infty$ and right-continuous at any point, there exists some $a$ s.t. $F_X(x)=0$ for $x<a$ and $F_X(x)=1$ for $x\geq a$. This is the CDF of degenerate distribution $\mathbb P(X=a)=1$.
For second point. If this is allowed, you can use the properties of variance.
$$
0=\text{Var}(X_1+\ldots+X_n)=\text{Var}(X_1)+\ldots+\text{Var}(X_n)\geq 0
$$
The r.h.s. is zero only if all the summands are zero. And $\text{Var}(X_i)=0$ implies that $X_i$ is a.s. constant.
EDIT: As Did pointed out, the equality between variance of the sum and te sum of variances requires square integrability of summands.
Prove the second point without additional requrements. Consider the case of two summands only.
Let $X$ and $Y$ are independent and $X+Y$ is a constant a.s. Prove that both $X,Y$ are a.s. constant.
By contradiction, let the distribution of $X$ is non-degenerate. Then there exists two disjoint intervals $[a,b]$ and $[c,d]$, where $b<c$ s.t.
$$\mathbb P(X\in[a,b])>0\quad \text{ and } \quad \mathbb P(X\in[c,d])>0.$$
Fix $\varepsilon>0$ s.t. $b+\varepsilon<c$. For $Y$, there exists some $z$ s.t.
$$\mathbb P(Y\in[z,z+\varepsilon])>0.$$
Then the intersection of independent events
$\{a\leq X \leq b\}$ and $\{z\leq Y\leq z+\varepsilon\}$ implies that
$$a+z \leq X+Y \leq b+z+\varepsilon < c+z,$$
and therefore
$$
\mathbb P(a+z \leq X+Y < c+z)\geq \mathbb P(a+z \leq X+Y\leq b+z+\varepsilon )\geq \mathbb P(a\leq X \leq b)\mathbb P(z\leq Y\leq z+\varepsilon)>0.
$$
Also
$$
\mathbb P(c+z \leq X+Y\leq d+z+\varepsilon )\geq \mathbb P(c\leq X \leq d)\mathbb P(z\leq Y\leq z+\varepsilon)>0.
$$
We prove that there exists two disjoint intervals $[a+z, c+z)$ and $[c+z,d+z+\varepsilon]$ s.t. probabilities of $X+Y$ belong to any of these intervals are positive, and then the distribution of $X+Y$ is non-degenerate too. Contradiction.
Best Answer
We have a sequence of i.i.d. (independent identically distributed) random variables $X_1,X_2,...$, where each $X_i\sim Exp(3)$. We know that their pdf is $f(x)=\lambda e^{-\lambda x}I_{(0,+\infty)} = 3e^{-3x}I_{(0,+\infty)}$. Consider the sequence of random variables: $$ \frac{\sum_{i=1}^n X_i + 3n}{\sum_{i=1}^n X_i^2} $$ Let's rewrite it as: $$ \frac{\sum_{i=1}^n X_i + 3n}{\sum_{i=1}^n X_i^2}\frac{n}{n} =\frac{n}{\sum_{i=1}^n X_i^2}\frac{\sum_{i=1}^n X_i + 3n}{n} = \frac{n}{\sum_{i=1}^n X_i^2}(\frac{\sum_{i=1}^n X_i}{n} + 3) $$ Let's focus on each term indipendently. By the Strong Law of Large Numbers, we know that: $$ \frac{\sum_{i=1}^n X_i^2}{n} \to E(X_i^2) = \frac{2}{\lambda^2} = \frac{2}{9} $$ $$ \frac{\sum_{i=1}^n X_i}{n} \to E(X_i) = \frac{1}{\lambda} = \frac{1}{3} $$ almost surely. Then, by using the continous mapping theorem, we can say that: $$ \frac{n}{\sum_{i=1}^n X_i^2}(\frac{\sum_{i=1}^n X_i}{n} + 3) \to \frac{9}{2}(\frac{1}{3}+3) = 15 $$ almost surely.