A contains 1.93 M hydrofluoric acid, HF (Ka = 7.2 × 10–4 ), and 3.00 M hydroponic acid, HCN (Ka = 6.2 × 10–10). What is the pH  ?

As Ka(HF) ≫ Ka(HCN), pH of the mixture almost entirely depends on HF.  The effect of HCN on pH is negligible.

             HF(aq) + H₂O(ℓ) ⇌ H₃O⁺(aq) + F⁻(aq)   Ka = 7.2 × 10⁻⁴
Initial:    1.93 M        --          0 M              0 M
Change:   -y  M        --        +y M            +y M
Eqm:   (1.93 - y) M   --         y M               y M
             ≈ 1.93 M

At eqm:
Ka(HF) = [H₃O⁺] [F⁻] / [HF]
7.2 × 10⁻⁴ = y² / 1.93
y = √(1.93 × 7.2 × 10⁻⁴) = 0.0373

pH = -log[H₃O⁺] = -log(0.0373) = 1.4

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OR:

pH
= -(1/2) {log(Ka) + log[HF]ₒ} 
= -(1/2) {log(7.2 × 10⁻⁴) + log(1.93)}
= 1.4