Here is a way to solve this problem using the convolution theorem:

l = Assuming[{\[Gamma] > 0 && \[Sigma] > 0 && \[Mu] > 0 && 
    k \[Element] Reals},
  FourierTransform[PDF[CauchyDistribution[\[Mu], \[Gamma]], x], x, k]
  ]

$\frac{\left(\theta (-k) e^{2 \gamma k}+\theta (k)\right) e^{-k (\gamma -i \mu )}}{\sqrt{2 \pi }}$

g = Assuming[{\[Gamma] > 0 && \[Sigma] > 0 && \[Mu] > 0 && 
    k \[Element] Reals},
  Expectation[Exp[I k x], 
   x \[Distributed] NormalDistribution[\[Mu], \[Sigma]]]
  ]

$e^{-\frac{k^2 \sigma ^2}{2}+i k \mu }$

Assuming[{\[Gamma] > 0 && \[Sigma] > 0 && \[Mu] > 0 && 
   k \[Element] Reals}, InverseFourierTransform[g l, k, x]]

$\frac{e^{-\frac{(-i \gamma -2 \mu +x)^2}{2 \sigma ^2}} \left(1-i \text{erfi}\left(\frac{-i \gamma -2 \mu +x}{\sqrt{2} \sigma }\right)\right)+e^{-\frac{(i \gamma -2 \mu +x)^2}{2 \sigma ^2}} \left(1+i \text{erfi}\left(\frac{i \gamma -2 \mu +x}{\sqrt{2} \sigma }\right)\right)}{2 \sqrt{2 \pi } \sigma }$

This is the result for the convolution as a function of x.

To take the Fourier transform of the Gaussian, I used Expectation with NormalDistribution (which is the Gaussian), applied to Exp[I k x]. That's faster than using FourierTransform directly - but the latter would also work.

Finally, I just take the inverse Fourier transform of the product of Fourier transforms, and fortunately we get an analytic result. It's still got some imaginary things in it, but they cancel. One could do some more work to get rid of them explicitly, but at least this should get you started.

By replacing the functions in your question with the built in distribution functions, I get slightly different prefactors. That can be amended by looking at the functional forms of the PDFs and comparing with your desired functions.

Here is a way to solve this problem using the convolution theorem:

l = Assuming[{γ > 0 && σ > 0 && μ > 0 && 
    k ∈ Reals},
  FourierTransform[PDF[CauchyDistribution[μ, γ], x], x, k]
  ]

$\frac{\left(\theta (-k) e^{2 \gamma k}+\theta (k)\right) e^{-k (\gamma -i \mu )}}{\sqrt{2 \pi }}$

g = Assuming[{γ > 0 && σ > 0 && μ > 0 && 
    k ∈ Reals},
  Expectation[Exp[I k x], 
   x \[Distributed] NormalDistribution[μ, σ]]
  ]

$e^{-\frac{k^2 \sigma ^2}{2}+i k \mu }$

Assuming[{γ > 0 && σ > 0 && μ > 0 && 
   k ∈ Reals}, InverseFourierTransform[g l, k, x]]

$\frac{e^{-\frac{(-i \gamma -2 \mu +x)^2}{2 \sigma ^2}} \left(1-i \text{erfi}\left(\frac{-i \gamma -2 \mu +x}{\sqrt{2} \sigma }\right)\right)+e^{-\frac{(i \gamma -2 \mu +x)^2}{2 \sigma ^2}} \left(1+i \text{erfi}\left(\frac{i \gamma -2 \mu +x}{\sqrt{2} \sigma }\right)\right)}{2 \sqrt{2 \pi } \sigma }$

This is the result for the convolution as a function of x.

To take the Fourier transform of the Gaussian, I used Expectation with NormalDistribution (which is the Gaussian), applied to Exp[I k x]. That's faster than using FourierTransform directly - but the latter would also work.

Finally, I just take the inverse Fourier transform of the product of Fourier transforms, and fortunately we get an analytic result. It's still got some imaginary things in it, but they cancel. One could do some more work to get rid of them explicitly, but at least this should get you started.

By replacing the functions in your question with the built in distribution functions, I get slightly different prefactors. That can be amended by looking at the functional forms of the PDFs and comparing with your desired functions.