The probability density function of an an exponential random variable with parameter $\lambda$ has probability density function
$$ f(x) = \begin{cases} \lambda e^{-\lambda x} & x \geq 0 \\ 0 & x < 0 \end{cases}$$
Therefore, the probability density function of $X$ is:
$$ f_X(x) = \begin{cases} e^{-x} & x \geq 0 \\ 0 & x < 0 \end{cases}$$
and the probability density function of $Y$ is:
$$ f_Y(y) = \begin{cases} 2e^{-2x} & x \geq 0 \\ 0 & x < 0 \end{cases}$$
The probability density function of $Z=X+Y$ is the convolution of the probability density functions of $X$ and $Y$. Therefore
\begin{align}
f_Z(z) &= \int_{-\infty}^{\infty} f_X(t)f_Y(z-t)\, dt
\end{align}
In order to get a formula for $f_Z$, we note that we care about when the integrand is nonzero. By the definitions of $f_X$ and $f_Y$, this is when $t \geq 0$ and $z \geq t$. Therefore, we care about integrating over the interval $0 \leq t \leq z$, noting that if $z$ is negative, $f_Z(z) = 0$. If such is the case, we reduce the integral to:
\begin{align}
f_Z(z) &= \int_{0}^{z} f_X(t)f_Y(z-t)\, dt \\
&= \int_0^{z} 2e^{-t}e^{2t-2z} \, dt \\
&= \int_0^{z} 2e^{t-2z} \, dt \\
&= 2e^{-2z} \int_0^{z} e^{t} \, dt \\
&= 2e^{-2z} (e^z - 1)
\end{align}
Therefore
$$ f_Z(z) = \begin{cases} 2e^{-2z} (e^z - 1) & z \geq 0 \\ 0 & x < 0 \end{cases}$$
Recall that by Bayes' Rule,
$$ f_{X|Z}(1,2) = \frac{f_{Z|X}(2,1) f_X(1)}{f_Z(2)} $$
The probability density function $f_{Z|X}$ is a measure of how likely $Z$ is to be equal to $z$ given that $X = x$. Since $Z = X + Y$, this is the same as measuring how likely it is that $Y$ will equal $z-x$. Therefore:
$$ f_{Z|X}(z,x) = f_Y(z-x) = \begin{cases} 2e^{-2(z-x)} & z \geq x \\ 0 & z < x \end{cases}$$
Therefore:
\begin{align}
f_{X|Z}(1,2) &= \frac{f_{Z|X}(2,1) f_X(1)}{f_Z(2)} \\
&= \frac{2e^{-2} \cdot e^{-1}}{2e^{-4}(e^2-1)} \\
&= \frac{e}{e^2-1}
\end{align}
