How to find frequency response, stability, and causality of a linear system?
NOTE: This answer is community wiki: If you feel there is a need for
  correction -please update the answer right here.


1. Frequency response:
For any system - the (discrete time or continuous time) Fourier Transform of the impulse response is same as the frequency response of it. You can replace $s$ by $j\omega$ to work on this.

2. Stability:
System is stable if, the bounded input produces bounded output. The definition of can be found here in the wiki link.  For this, the necessary condition to prove is to see if the impulse response be absolutely integrable, i.e., its L^1 norm exist.

For a rational and continuous-time system, the condition for stability is that the region of convergence (ROC) of the Laplace transform includes the imaginary axis.

3. Causality:
Any system who's impulse response at sample $s_i$ doesn't require to know any samples presented $s_{i+1}$ or after wards, than the system is causal.

EDIT:
Now given the above i am trying to put the calculation as per your system.

$$ H(s)=\frac{s}{(s+1)(s+2)}  $$

Now we can replace $s$ with $j\omega$ to get the frequency response in terms of magnitude and phase response. 

$$ H(j\omega) =  \frac{j\omega}{(j\omega+1)(j\omega+2)} $$
$$ H(j\omega) =  \frac{j\omega}{2 - \omega^2 + 3j\omega } $$
$$ Magnitude =  20 Log_{10} \sqrt { ( H(j\omega) )^2 } $$
$$ Magnitude =  20 Log_{10}  \sqrt { \frac { (j\omega)^2 }{(2 - \omega^2)^2 + (3j\omega)^2} }  $$

NOTE: square is applied separately on real terms than on imaginary terms! I got this understanding from this document here. 

Replacing $-j^2$ by $-1$

$$ Magnitude =  20 Log_{10} \sqrt { \frac{-1\cdot\omega^2}{(2 - \omega^2)^2 - (3\omega)^2} } $$

$$ Magnitude =  20 Log_{10} \sqrt { \frac{-1\cdot\omega^2}{(4 - 5\omega^2 +\omega^4)} } $$

$$ Phase = arc tan (real / imaginary)  $$

$$ Phase = -1 \cdot \tan^{-1}(\frac {-3\omega}{(2-\omega^2)} )   \ $$ 

Please correct me if i am wrong. 

Another literature that can help you.