Yes. That's correct. A PDF is a probability density function. It is stating the probability of a particular value coming out. Taking this analogy to a discrete distribution, the PDF of a 6-sided die is: $[x<1:0,x=1:\frac{1}{6},x=2:\frac{1}{6},x=3:\frac{1}{6},x=4:\frac{1}{6},x=5:\frac{1}{6},x=6:\frac{1}{6},x>6:0]$. For a continuous probability distribution, you can't really use the PDF directly, since the probability of an infinitesimally thin slice of the PDF being selected is intuitively zero.
That's where the cumulative density function, or CDF, comes it. It is a measure of how likely the value is to be less than some arbitrary value (which we pick). For a discrete case, you start with the first possible value, and add all the entries in the PDF up to the value of interest:
$$CDF=\sum PDF \rightarrow [x<1:0,x<2:\frac{1}{6},x<3:\frac{2}{6},x<4:\frac{3}{6},x<5:\frac{4}{6},x<6:\frac{5}{6},x\geq 6:\frac{6}{6}]$$
Notice how the final value of the CDF is $1$. This is expected, since every possible outcome of rolling a 6-sided die is less than or equal to 6.
Now let's go back to the continuous probability distribution. In this case, we don't have a finite set of options for the answer to be, so we can't constrain $X$. Thus, we start from $-\infty$, since that encompasses everything to the left of the chosen $x$. As you should be aware from calculus, the integral is to continuous functions what a sum is to discrete functions - loosely. The value of a CDF is that you can use it to determine the probability of the number falling within a specific range as follows:
$$F(a\leq X \leq b) = F(X \leq b) - F(X \leq a) = \int_{-\infty}^{b} f(x)dx - \int_{-\infty}^{a} f(x)dx = \int_{a}^{b} f(x)dx$$
$V$ is a function of the random variable $T$, and the behavior of $V$ changes when $T=3$. So to calculate the probability $P(V\le v)$ you should argue differently depending on whether $T\le3$ or $T>3$. This means you break up the probability into two pieces:
$$P(V\le v)=P(V\le v, T\le3) + P(V\le v, T>3)\tag1$$
To decide how to proceed from here, look at what values $V$ can take: It has a mass at value $5$, and values from $6$ onward. So to cover all possibilities, you should consider four cases: (a) $v<5$ (b) $v=5$ (c) $5< v<6$ (d) $v \ge 6$.
For case (a), the second term on the RHS of (1) is zero because $V>6$ when $T>3$ so it's impossible for $V\le v$. The first term is also zero because $V=5$ when $T\le3$.
For case (b) and (c), the second term is zero using the same argument as in (a). You should get the first term to be $P(T\le3)$, because $V=5$ when $T\le3$, so for sure $V\le v$.
For case (d), the first term on the RHS of (1) simplifies to $P(T\le3)$ using the same reasoning as in the previous case. The second term simplifies to $P(2T\le v,T>3)$ because $V=2T$ when $T>3$. Therefore (1) becomes
$$P(V\le v)=P(T\le3)+P(3<T\le v/2)=P(T\le v/2).$$
Best Answer
"Proportional to $x$" means that it is of the form $f(x)=cx$ for some constant $c$.
Now, for it to be a probability density, it needs to integrate to 1, so $$\int_\mathbb Rf(x)\,dx = \int_0^1 cx\,dx = c \left[\frac{x^2}{2}\right]_0^1 = c\left(\frac{1}{2}-\frac{0}{2}\right)=\frac{c}{2}=1.$$
From this you can figure out the value of $c$ that makes the function be a probability density.
EDIT: once you have the density function, you can also find the cumulative distribution function by integration: for every $x$ in $[0,1]$, $$F(x)=\int_{-\infty}^x f(t)\,dt=\int_0^x 2t\,dt = 2\left[\frac{t^2}{2}\right]_0^x=(x^2-0)=x^2.$$
You can check it fulfills all the conditions to be a cumulative distribution: it is always nonnegative, when $x\rightarrow -\infty$ it tends to $0$ (because $F(0)=0$), when $x\rightarrow +\infty$ it tends to $1$ (because $F(1)=1$), and it is right-continuous (it is actually continuous).