Yes, the fact that you know that $\Pr(X\gt x)=e^{-\lambda x}$ (if $x\ge 0$) is very useful. Equivalently, $F(x)=1-e^{-\lambda x}$ when $x\ge 0$, and $F(x)=0$ elsewhere. Being aware of the "elsewhere" part can help avoid error.
For the first problem, just substitute $2/\lambda$ for $x$. Note that $\lambda(2/\lambda)=2$.
For the second, let's "unwrap" what $|X-1/\lambda|\lt 2/\lambda$ says. It says that $X$ is not too far from $1/\lambda$, indeed within $2/\lambda$ of $1/\lambda$. More precisely, it says that $\cfrac{1}{\lambda}-\cfrac{2}{\lambda} \lt X \lt \cfrac{1}{\lambda}+\cfrac{2}{\lambda}.$ Some simplification reduces this to $-\cfrac{1}{\lambda}\lt X\lt \cfrac{3}{\lambda}.$ But note that an exponentially distributed random variable can never be negative. So all we want is $\Pr\left(X\lt \cfrac{3}{\lambda}\right).$ Another way of arriving at the answer is to note that our probability is $F(3/\lambda)-F(-1/\lambda)$. But now we must be careful. Since $-1/\lambda$ is negative, $F(-1/\lambda)=0$, so our answer is just $F(3/\lambda)$. We get this by substituting $3/\lambda$ for $x$ in the formula for $F(x)$.